题1：
Suppose a sorted array 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).
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];}
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