33. Search in Rotated Sorted Array

题目

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.

分析

老题了，就不多写了，二分+注意边界条件，如果不确定边界条件把所有的情况都列出来，再将相同处理的情况进行融合。

代码

public int search(int[] nums, int target) {
    if(nums == null || nums.length == 0) return -1;
    int left = 0, right = nums.length - 1;
    while(left < right){
        int mid = left + (right - left) / 2;
        if(nums[mid] == target){
            return mid;
        }
        if(nums[mid] >= nums[left]){ //左侧有序
            if(nums[mid] > target && nums[left] <= target){
                right = mid - 1;
            }else{
                left = mid + 1;
            }
        }else{  //右侧有序
            if(nums[mid] <= target && nums[right] >= target){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }
    }
    return nums[left] == target ? left : -1;
}