[LeetCode]153 有序数组旋转后的最小值

Find Minimum in Rotated Sorted Array(找到有序数组旋转后的最小值)

【难度：Medium】
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.
假设一有序数组绕某点旋转，找到其中最小的值，数组中无重复值出现。

---

解题思路

本题可使用常规遍历查找或是二分查找来解决。
当使用二分查找时，要判断的是中间值nums[mid]与末端值nums[high]的大小比较，

• 1)若nums[mid]>nums[high]，虽然数组经过旋转之后顺序改变了，但是可以确定最小值出现在[mid+1,high]这部分区间之中;
• 2)否则出现在[low,mid]这部分区间之中。

---

c++代码如下：

//二分查找class Solution {public:    int findMin(vector<int>& nums) {        if (nums.empty())            return 0;        int low = 0;        int high = nums.size()-1;        int mid = 0;        while (low < high) {            mid = (low+high)/2;            nums[mid] > nums[high] ? low = mid+1 : high = mid;        }        return nums[high];    }};

//常规查找class Solution {public:    int findMin(vector<int>& nums) {        if (nums.empty())            return 0;        int min = nums[0];        for(int i = 1; i < nums.size(); i++) {            if (nums[i] < min)                min = nums[i];        }        return min;    }};