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lucifer
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feat:0.6.0
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-4
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‎leetcode-cheat-0.5.1.zip

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‎leetcode-cheat-0.6.0.zip

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‎public/manifest.json

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@@ -2,7 +2,7 @@
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"manifest_version": 2,
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"name": "leetcode cheatsheet",
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"description": "刷题小助手，made by 力扣加加",
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"version": "0.5.2",
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"version": "0.6.0",
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"browser_action": {
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"default_popup": "index.html",
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"default_title": "力扣加加"

‎src/db/root.db.js

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Original file line numberDiff line numberDiff line change
@@ -3959,13 +3959,21 @@
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"giteeSolution": "https://gitee.com/golong/leetcode/blob/master/problems/128.longest-consecutive-sequence.md",
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"solution": "https://github.com/azl397985856/leetcode/blob/master/problems/128.longest-consecutive-sequence.md",
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"code": [
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{
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"language": "java",
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"text": "\nclass Solution {\n public int longestConsecutive(int[] nums) {\n Set<Integer> set = new HashSet<Integer>();\n int ans = 0;\n for (int num : nums) {\n set.add(num);\n }\n for(int i = 0;i < nums.length; i ++) {\n int x = nums[i];\n // 说明x是连续序列的开头元素\n if (!set.contains(x - 1)) {\n while(set.contains(x + 1)) {\n x ++;\n }\n }\n ans = Math.max(ans, x - nums[i] + 1);\n }\n return ans;\n \n }\n}\n"
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},
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{
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"language": "js",
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"text": "\nif (nums.length === 0) return 0;\nlet count = 1;\nlet maxCount = 1;\n// 这里其实可以不需要排序，这么做只不过是为了方便理解\nnums = [...new Set(nums)].sort((a, b) => a - b);\nfor (let i = 0; i < nums.length - 1; i++) {\n if (nums[i + 1] - nums[i] === 1) {\n count++;\n } else {\n if (count > maxCount) {\n maxCount = count;\n }\n count = 1;\n }\n}\nreturn Math.max(count, maxCount);\n"
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},
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{
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"language": "js",
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"text": "\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar longestConsecutive = function(nums) {\n set = new Set(nums);\n let max = 0;\n let temp = 0;\n set.forEach(x => {\n // 说明x是连续序列的开头元素\n if (!set.has(x - 1)) {\n temp = x + 1;\n while (set.has(y)) {\n temp = temp + 1;\n }\n max = Math.max(max, y - x); // y - x 就是从x开始到最后有多少连续的数字\n }\n });\n return max;\n};\n"
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"text": "\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar longestConsecutive = function(nums) {\n set = new Set(nums);\n let max = 0;\n let temp = 0;\n set.forEach(x => {\n // 说明x是连续序列的开头元素。加这个条件相当于剪枝的作用，否则时间复杂度会退化到 N ^ 2\n if (!set.has(x - 1)) {\n temp = x + 1;\n while (set.has(y)) {\n temp = temp + 1;\n }\n max = Math.max(max, y - x); // y - x 就是从x开始到最后有多少连续的数字\n }\n });\n return max;\n};\n"
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},
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{
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"language": "py",
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"text": "\nclass Solution:\n def longestConsecutive(self, A: List[int]) -> int:\n seen = set(A)\n ans = 0\n for a in A:\n t = a\n # if 的作用是剪枝\n if t + 1 not in seen:\n while t - 1 in seen:\n t -= 1\n ans = max(ans, a - t + 1)\n return ans\n"
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}
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]
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},
@@ -6728,6 +6736,42 @@
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}
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]
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},
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"patching-array":{
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"id": "330",
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"name": "patching-array",
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"pre": [
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{
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"text": "贪心",
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"link": null,
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"color": "purple"
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},
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{
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"text": "前缀和",
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"link": null,
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"color": "cyan"
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}
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],
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"keyPoints": [
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{
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"text": "维护端点信息，并用前缀和更新",
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"link": null,
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"color": "blue"
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}
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],
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"companies": [],
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"giteeSolution": "https://gitee.com/golong/leetcode/blob/master/problems/330.patching-array.md",
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"solution": "https://github.com/azl397985856/leetcode/blob/master/problems/330.patching-array.md",
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"code": [
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{
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"language": "py",
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"text": "\nclass Solution:\n def minPatches(self, nums: List[int], n: int) -> int:\n furthest = i = ans = 0\n while furthest < n:\n # 可覆盖到，直接用前缀和更新区间\n if i < len(nums) and nums[i] <= furthest + 1:\n furthest += nums[i] # [1, furthest] -> [1, furthest + nums[i]]\n i += 1\n else:\n # 不可覆盖到，增加一个数 furthest + 1，并用前缀和更新区间\n furthest = 2 * furthest + 1 # [1, furthest] -> [1, furthest + furthest + 1]\n ans += 1\n return ans\n\n"
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},
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{
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"language": "py",
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"text": "\nclass Solution:\n def minPatches(self, nums: List[int], n: int) -> int:\n furthest, i, ans = 1, 0, 0\n # 结束条件也要相应改变\n while furthest <= n:\n if i < len(nums) and nums[i] <= furthest:\n furthest += nums[i] # [1, furthest) -> [1, furthest + nums[i])\n i += 1\n else:\n furthest = 2 * furthest # [1, furthest) -> [1, furthest + furthest)\n ans += 1\n return ans\n"
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}
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]
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},
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"increasing-triplet-subsequence":{
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"id": "334",
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"name": "increasing-triplet-subsequence",
@@ -6758,7 +6802,7 @@
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"code": [
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{
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"language": "js",
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"text": "\n\n\n/*\n * @lc app=leetcode id=334 lang=javascript\n *\n * [334] Increasing Triplet Subsequence\n *\n * https://leetcode.com/problems/increasing-triplet-subsequence/description/\n *\n * algorithms\n * Medium (39.47%)\n * Total Accepted: 89.6K\n * Total Submissions: 226.6K\n * Testcase Example: '[1,2,3,4,5]'\n *\n * Given an unsorted array return whether an increasing subsequence of length 3\n * exists or not in the array.\n * \n * Formally the function should:\n * \n * Return true if there exists i, j, k \n * such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return\n * false.\n * \n * Note: Your algorithm should run in O(n) time complexity and O(1) space\n * complexity.\n * \n * \n * Example 1:\n * \n * \n * Input: [1,2,3,4,5]\n * Output: true\n * \n * \n *\n * Example 2:\n * \n * \n * Input: [5,4,3,2,1]\n * Output: false\n * \n * \n * \n */\n/**\n * @param {number[]} nums\n * @return {boolean}\n */\nvar increasingTriplet = function(nums) {\n if (nums.length < 3) return false;\n let n1 = Number.MAX_VALUE;\n let n2 = Number.MAX_VALUE;\n\n for(let i = 0; i < nums.length; i++) {\n if (nums[i] <= n1) {\n n1 = nums[i]\n } else if (nums[i] <= n2) {\n n2 = nums[i]\n } else {\n return true;\n }\n }\n\n return false;\n};\n"
6805+
"text": "\n/*\n/**\n * @param {number[]} nums\n * @return {boolean}\n */\nvar increasingTriplet = function(nums) {\n if (nums.length < 3) return false;\n let n1 = Number.MAX_VALUE;\n let n2 = Number.MAX_VALUE;\n\n for(let i = 0; i < nums.length; i++) {\n if (nums[i] <= n1) {\n n1 = nums[i]\n } else if (nums[i] <= n2) {\n n2 = nums[i]\n } else {\n return true;\n }\n }\n\n return false;\n};\n"
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}
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]
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},
@@ -7255,6 +7299,57 @@
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}
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]
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},
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"binary-watch":{
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"id": "401",
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"name": "binary-watch",
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"pre": [
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{
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"text": "笛卡尔积",
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"link": null,
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"color": "red"
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},
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{
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"text": "回溯",
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"link": "https://github.com/azl397985856/leetcode/blob/master/thinkings/backtrack.md",
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"color": "green"
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}
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],
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"keyPoints": [],
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"companies": [
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{
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"name": "阿里巴巴"
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},
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{
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"name": "腾讯"
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},
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{
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"name": "百度"
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},
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{
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"name": "字节跳动"
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}
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],
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"giteeSolution": "https://gitee.com/golong/leetcode/blob/master/problems/401.binary-watch.md",
7333+
"solution": "https://github.com/azl397985856/leetcode/blob/master/problems/401.binary-watch.md",
7334+
"code": [
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{
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"language": "py",
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"text": "\n# 枚举小时\nfor a in possible_number(i):\n # 小时确定了，分就是 num - i\n for b in possible_number(num - i, True):\n ans.add(str(a) + \":\" + str(b).rjust(2, '0'))\n"
7338+
},
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{
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"language": "py",
7341+
"text": "\nfor i in range(min(4, num + 1)):\n for a in possible_number(i):\n for b in possible_number(num - i, True):\n ans.add(str(a) + \":\" + str(b).rjust(2, '0'))\n"
7342+
},
7343+
{
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"language": "py",
7345+
"text": "\nclass Solution:\n def readBinaryWatch(self, num: int) -> List[str]:\n def possible_number(count, minute=False):\n if count == 0: return [0]\n if minute:\n return filter(lambda a: a < 60, map(sum, combinations([1, 2, 4, 8, 16, 32], count)))\n return filter(lambda a: a < 12, map(sum, combinations([1, 2, 4, 8], count)))\n ans = set()\n for i in range(min(4, num + 1)):\n for a in possible_number(i):\n for b in possible_number(num - i, True):\n ans.add(str(a) + \":\" + str(b).rjust(2, '0'))\n return list(ans)\n"
7346+
},
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{
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"language": "py",
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"text": "\nclass Solution:\n def readBinaryWatch(self, num: int) -> List[str]:\n return [str(a) + \":\" + str(b).rjust(2, '0') for a in range(12) for b in range(60) if (bin(a)+bin(b)).count('1') == num]\n"
7350+
}
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]
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},
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"partition-equal-subset-sum":{
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"id": "416",
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"name": "partition-equal-subset-sum",
@@ -7631,7 +7726,7 @@
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},
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{
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"language": "py",
7634-
"text": "\nclass Solution:\n def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:\n # acc 表示当前累计的数字和\n def dfs(acc):\n if acc >= desiredTotal:\n return False\n for n in range(1, maxChoosableInteger + 1):\n # 对方有一种情况赢不了，我就选这个数字就能赢了，返回 true，代表可以赢。\n if not backtrack(acc + n):\n return True\n return False\n\n # 初始化集合，用于保存当前已经选择过的数。\n return dfs(0)\n"
7729+
"text": "\nclass Solution:\n def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:\n # acc 表示当前累计的数字和\n def dfs(acc):\n if acc >= desiredTotal:\n return False\n for n in range(1, maxChoosableInteger + 1):\n # 对方有一种情况赢不了，我就选这个数字就能赢了，返回 true，代表可以赢。\n if not dfs(acc + n):\n return True\n return False\n\n # 初始化集合，用于保存当前已经选择过的数。\n return dfs(0)\n"
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},
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{
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"language": "py",
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}
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]
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},
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"jump-game-iv":{
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"id": "1435",
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"name": "jump-game-iv",
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"pre": [
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{
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"text": "BFS",
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"link": null,
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"color": "purple"
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}
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],
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"keyPoints": [],
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"companies": [],
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"giteeSolution": "https://gitee.com/golong/leetcode/blob/master/problems/1435.jump-game-iv.md",
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"solution": "https://github.com/azl397985856/leetcode/blob/master/problems/1435.jump-game-iv.md",
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"code": [
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{
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"language": "py",
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"text": "\nclass Solution:\n def minJumps(self, A: List[int]) -> int:\n dic = collections.defaultdict(list)\n n = len(A)\n\n for i, a in enumerate(A):\n dic[a].append(i)\n visited = set([0])\n q = collections.deque([0])\n steps = 0\n\n while q:\n for _ in range(len(q)):\n i = q.popleft()\n visited.add(i)\n if i == n - 1: return steps\n for neibor in dic[A[i]] + [i - 1, i + 1]:\n if 0 <= neibor < n and neibor not in visited:\n q.append(neibor)\n # 剪枝\n dic[A[i]] = []\n steps += 1\n return -1\n"
10221+
}
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]
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},
1010810224
"form-largest-integer-with-digits-that-add-up-to-target":{
1010910225
"id": "1449",
1011010226
"name": "form-largest-integer-with-digits-that-add-up-to-target",

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