33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

题意：循环有序序列，前半段是有序的，后半段也是有序的。找到目标节点的序号。

class Solution {    public int search(int[] nums, int target) {        int left=0,right=nums.length-1;        while(left<=right){            int mid=(left+right)/2;            if (nums[mid]==target) return mid;            if (nums[mid]>=nums[left]){                if(target<nums[mid]&&target>=nums[left]) right=mid-1;                else left=mid+1;            }else{                if (target>nums[mid]&&target<=nums[right]) left=mid+1;                else right=mid-1;            }        }        return -1;    }}

1.在nums[mid]>=nums[left]，左半段是有序的。

先在做半段找target，首先nums[left]<=target<nums[mid]，保证target在做半段中，right=mid-1;

2.在右半段是有序的，nums[mid]<target<=nums[right]，left=mid+1;