Description

Suppose a sorted array 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).

Find the minimum element.

Notice

You may assume no duplicate exists in the array.

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

4/24/2017

算法班

不需要跟特定值比较，while里面判断只有2个分支。

1 public class Solution { 2     /** 3      * @param nums: a rotated sorted array 4      * @return: the minimum number in the array 5      */ 6     public int findMin(int[] nums) { 7         // write your code here 8         if (nums == null || nums.length == 0) return -1; 9         int start = 0, end = nums.length - 1;10         11         while (start + 1 < end) {12             int mid = start + (end - start) / 2;13             if (nums[mid] < nums[end]) {14                 end = mid;15             } else {16                 start = mid;17             }18         }19         return nums[start] < nums[end]? nums[start]: nums[end];20     }21 }