每日一题DAY8寻找旋转排序数组最小值-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43637099/article/details/115534345

//升序的话做旋转数组，划一刀两边至少一边有序
//左边有序，排列为k…mid…max 0 …k-1，
//右边有序，排序为k …max 0 … mid … k-1 (注意mid = 0)
//注意left可能=mid,right一定>mid，为避免=时的情况考虑是否漏掉mid，改用right判断

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

整不明白非要用left的话自己写就知道容易出错了
用left的话可以加个res保留结果

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0,right = nums.size()-1;
        int res = nums[0];
        while(left <= right){
            int mid = left + ((right-left)>>1);
            res = min(res,nums[mid]);//比较中间值
            if (nums[left]<=nums[mid]){//左边有序
                res = min(res,nums[left]);//直接得到左边最小值
                left = mid+1;//搜索右边
            }
            else{
                res = min(res,nums[mid+1]);//直接得到右边最小值
                right = mid - 1;//搜索左边
            }
        }
        return res;
    }
};