LeetCode-Find Minimum in Rotated Sorted Array（找出反转数组中的最小值）

题1：

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.

You may assume no duplicate exists in the array.

题2：如果允许上题的数组有重复数字呢？

解法：最笨的方法，也是最容易想到的方法就是直接遍历数组，找出最小值，时间复杂度O(n)，但是对于本题明显会超时。此时，比较容易想到的就是二分查找，因为其有序。但是两者肯定是有不同点的。

第一题的代码如下：

public int findMin(int[] num) {        int low = 0;        int high = num.length-1;        while (low < high-1) {        int mid = low + (high-low)/2;        if (num[low] < num[high]) {        if (num[mid] < num[low]) {        low = mid;        } else {        high = mid;        }        } else {        if (num[mid] < num[high]) {        high = mid;        } else {        low = mid;        }        }        }        return num[low] > num[high] ? num[high] : num[low];    }

此题一定要区分，这个序列是递增的还是递减的。很多解题方法都没考虑这个问题。显然有了第一题的思路，第二题也就好解了。

public int findMin(int[] num) {        int low = 0;        int high = num.length-1;        while (low < high-1) {        int mid = low + (high-low)/2;        if (num[low] < num[high]) {        if (num[mid] < num[low]) {        low = mid;        } else if (num[mid] > num[low]){        high = mid;        } else {        low++;        }        } else {        if (num[mid] < num[high]) {        high = mid;        } else if (num[mid] > num[high]){        low = mid;        } else {        high--;        }        }        }        return num[low] > num[high] ? num[high] : num[low];    }