86. Partition List

• Total Accepted: 89180
• Total Submissions: 281459
• Difficulty: Medium
• Contributors: Admin

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

来源： https://leetcode.com/problems/partition-list/

分析

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使用两个数组存储小于 x 和 不小于 x的数字，然后重新幅值给list

时间复杂度O(N), 空间复杂度O(N)

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/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public:     ListNode* partition(ListNode* head, int x) {         /** we can use two vector<int> store different types of numbers, n1 record the number smaller than x, n2 record           *  the numbers no-smaller than x          */          if(head == NULL || head->next == NULL) return head;                     vector<int> n1, n2;          ListNode* p = head;          while(p != NULL){              p->val < x ? n1.push_back(p->val) : n2.push_back(p->val);              p = p->next;          }          p = head;          for(auto n: n1){              p->val = n;              p = p->next;          }          for(auto n: n2){              p->val = n;              p = p->next;          }                     return head;               } };

新建两个 node ,分别存储小于 和 不小于的数字，然后合并

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/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public:     ListNode* partition(ListNode* head, int x) {         if(head == NULL || head->next == NULL) return head;                   ListNode dummy1(0), dummy2(0);         ListNode* p1 = &dummy1, * p2 = &dummy2;         while(head != NULL){             // using reference to reduce the code lines             ListNode* & ref = head->val < x ? p1 : p2;             ref->next = head;             ref = ref->next;             /**             if(head->val < x){                 p1->next = head;                 p1 = p1->next;             }             else{                 p2->next = head;                 p2 = p2->next;             }             */             head = head->next;         }         p1->next = dummy2.next;                   // notice that this step makes sure that there is no loop in the new list         p2->next = NULL;         return dummy1.next;     } };

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