[leetcode]partition-list

将list中比x小的node都放左边， 比x大的都放右边

思路

一次遍历，遇小前叉，关键是要维护一个指向要插入位置的node*

Solution 1

/**
 * 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) {
        ListNode* hair = new ListNode(0);
        hair->next = head;
        ListNode* insertPoint = hair; //insert after this pointer
        ListNode* last = hair;
        while(head){
            if(head->val < x){
                if(insertPoint->next != head){//要是没有这句，如果第一个就比x小，会先delete再insert回原样
                    //backup 
                    ListNode* nextNode = head->next;

                    //do delete , last不用动
                    last->next = head->next;

                    //do insert
                    ListNode* temp = insertPoint->next;
                    insertPoint->next = head;
                    head->next = temp;

                    //update head
                    insertPoint = head;
                    head = nextNode; 
                    //last不用动

                }else{ //insertPoint->next == head
                    //update insert point
                    insertPoint = head;
                    head = head->next;
                    last = last->next;
                }
            }else{
                head = head->next;  //head 一定比insertPoint跑的远，因为本身就起点靠前，跑的又快
                last = last->next;
            }

        }
        return hair->next;
    }
};