Find Minimum in Rotated Sorted Array 解题报告

Find Minimum in Rotated Sorted Array

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).

Notice
You may assume no duplicate exists in the array.

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

实现思路

1. 使用二分法实现log(n)时间复杂度查找
2. 注意好mid的求法，直接用(right + left) / 2求可能越界
3. 思考判断条件，如果中间值大于右边，说明中间值属于折到前面的较大数，要查找最小数，故往右边找，且left = mid + 1;,如果中间值小于右边，说明中间值属于原来较小的数，说明最小值在左边或此时位置恰好为最小值，故right = mid。

/** * @param nums: a rotated sorted array * @return: the minimum number in the array */public int findMin(int[] nums) {    // write your code here    int len = nums.length;    if(len == 1){        return nums[0];    }    int left = 0,right = len-1;    int min = nums[0];    while(left < right){        int mid = (right - left) / 2 + left;        if(nums[mid] > nums[right]){            left = mid + 1;        }else{            right = mid;        }    }    return nums[left];}