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Leetcode--35. 搜索插入位置

忆北文学摄影爱好员 2022-04-17 阅读 82
算法c++

题目链接:35. 搜索插入位置

代码1:

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

代码2:

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

mid=(left+right)/2

存在问题。

原因:left可能不断增大,如果到极限状态,也就是left达到了right-1的地步的时候刚好数组的长度又很大,那么就可能导致left + right的溢出出现负数;

mid=left+(right-left)/2或mid=left+((right-left)>>2)

(right+left)相加的结果可能会导致整型溢出的情况,(right-left)使用减法不会超出最大的整型范畴;

>>是右移运算符,右移一位相当于除2,右移n位相当于除以2的n次方;mid=(left+right)>>1等价于mid=(left+right)/2;

left+(right-left)/2通分可知是等同于 (left + right) / 2;

 

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