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partition_list.cpp
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/**
* @Author: Chacha
* @Date: 2019-01-10 20:22:11
* @Last Modified by: Chacha
* @Last Modified time: 2019-01-10 21:45:48
*/
#include<iostream>
#include<string>
using namespace std;
/**
* Definition for singly-linked list(单向链表)
* Source: https://zh.wikipedia.org/wiki/链表
*/
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
/**
* Given a linked list and a value x, partition it such that
* all nodes less than x come before nodes greater than or equal to x.
* You should preserve the original relative order of the nodes in each of the two partitions.
*
* Example:
* Input: head = 1->4->3->2->5->2, x = 3
* Output: 1->2->2->4->3->5
*
* Source: https://leetcode.com/problems/partition-list/
*
*/
ListNode* partitionList(ListNode* head, int x) {
if (head == NULL) return NULL;
ListNode* leftDummy = new ListNode(0);
ListNode* rightDummy = new ListNode(0);
ListNode* left = leftDummy;
ListNode* right = rightDummy;
while(head != NULL){
if (head->val < x) {
left->next = head;
left = left->next;
} else {
right->next = head;
right = right->next;
}
head = head->next;
}
right->next = NULL;
left->next = rightDummy->next;
return leftDummy;
}
};
/* Function to print nodes in a given linked list */
void printList(ListNode *node) {
while (node != NULL) {
printf("%d ", node->val);
node = node->next;
}
}
int main() {
ListNode* head = new ListNode(1);
head->next = new ListNode(4);
head->next->next = new ListNode(3);
head->next->next->next = new ListNode(2);
head->next->next->next->next = new ListNode(5);
head->next->next->next->next->next = new ListNode(2);
ListNode* result = Solution().partitionList(head, 3);
printList(result);
return 0;
}