Create 153. Find Minimum in Rotated Sorted Array.md

solutions/153. Find Minimum in Rotated Sorted Array.md

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+# [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+# 思路
+要求一个经过循环移位后的排序数组(无重复元素)中的最小值. 
+其实这题是[33. Search in Rotated Sorted Array](https://github.com/ShusenTang/LeetCode/blob/master/solutions/33.%20Search%20in%20Rotated%20Sorted%20Array.md)
+的前半部分, 所以搞懂33题了这题肯定完全没问题.
+
+由于数组(移位前)是有序的所以肯定是二分, 设最小值在i处, 则数组A = nums[0,...,i-1]和B = nums[i,...,n-1]是分别递增的, 而且B的所有元素都小于A任意元素.
+二分while循环中mid需要分三种情况:
+1. 当`nums[low] > nums[mid]`时, 说明mid在B部分, 则i在区间[low, mid];
+2. 否则即mid在A部分(B存不存在未知), 此时当`nums[low] > nums[high]`时, 说明B部分存在, 则i在区间[mid+1, high];
+3. 否则, 即mid在A部分且B不存在, 即low到high是严格递增的, 最小值即nums[low], 直接返回即可.
+
+> 注意边界情况.
+
+# C++
+``` C++
+class Solution {
+public:
+    int findMin(vector<int>& nums) {
+        int low = 0, high = nums.size() - 1, mid;
+        
+        while(low <= high){
+            mid = low + (high - low) / 2;
+            if(nums[low] > nums[mid]) high = mid;
+            else if(nums[low] > nums[high]) low = mid + 1;
+            else return nums[low];
+        }
+        return nums[mid];
+    }
+};
+```