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 7might become4 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

先判断m在上半段还是下半段，在每个上半段里，判断target是在m前还是后，m前就是是递增区间，可以直接确定范围，如果是m后，则回到初始问题一样了，继续判断。

class Solution {
    public int search(int[] nums, int target) {
        if(nums == null || nums.length == 0){
            return -1;
        }
        int s = 0, e = nums.length - 1;

        while(s + 1 < e){
            int m = s + (e - s)/2;

            if(nums[s] < nums[m]){
              //m在上半段
                //判断target是在m前还是后，m前是递增区间，可以直接确定范围，如果是m后，则和初始问题一样了，继续判断
                if(nums[s] <= target && target <= nums[m]){
                    e = m;//递增区间
                }else{
                    s = m;//回到原题
                }
            }else{
                //m在下半段
                if(nums[m] <= target && target <= nums[e]){
                    s = m;
                }else{
                    e = m;
                }
            }

        }

        if(nums[s] == target){
            return s;
        }
        if(nums[e] == target){
            return e;
        }  
          return -1;  
    }
}