**Leetcode 153. Find Minimum in Rotated Sorted Array

Posted 2023-01-27 Z-Pilgrim

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/

面试最常见的题之一了

补充一些细节问题：

1）为什么不能根据mid和nums[0]来做二分（如果nums[mid] > nums[0] 在mid右边找）

2）为什么要mid对比右边界

对于1）很容易举出来反例，如果数组是递增的，在mid右边找自然是不对的

                特判nums[0]和nums[length-1]能特判递增情况

        2）递增是一种bad case

                非递增一个bad case   [4,5,6,7,0,1,2] ，由于数组本身最小值是在子数组的左边，当做完一轮二分之后，有可能子数组是递增的，这个时候，继续对比nums[l] 和nums[mid] 就会出问题

class Solution 
public:
    int findMin(vector<int>& nums) 
        int left = 0, right = nums.size() - 1;
        int mid = 0, en = right;
        while (left <= right) 
            mid = (left + right)/2;
            if (mid && nums[mid] < nums[mid-1]) return nums[mid];
            
            if (nums[mid] > nums[right]) 
                left = mid + 1;
             else 
                right = mid - 1;
            
        
        return nums[mid];
    
;

其他写法：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/discuss/48484/A-concise-solution-with-proof-in-the-comment

class Solution 
public:
    int findMin(vector<int> &num) 
        int low = 0, high = num.size() - 1;
        // loop invariant: 1. low < high
        //                 2. mid != high and thus A[mid] != A[high] (no duplicate exists)
        //                 3. minimum is between [low, high]
        // The proof that the loop will exit: after each iteration either the 'high' decreases
        // or the 'low' increases, so the interval [low, high] will always shrink.
        while (low < high) 
            auto mid = low + (high - low) / 2;
            if (num[mid] < num[high])
                // the mininum is in the left part
                high = mid;
            else if (num[mid] > num[high])
                // the mininum is in the right part
                low = mid + 1;
        

        return num[low];
    
;