每日算法——leetcode系列

問題 3Sum Closest

Difficulty: Medium

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

<pre>
For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

</pre>

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        
    }
};

翻譯

三數(shù)和

難度系數(shù)：中等

給定一個有n個整數(shù)的數(shù)組S, 找出數(shù)組S中三個數(shù)的和最接近一個指定的數(shù)。 返回這三個數(shù)的和。 你可以假定每個輸入都有一個結(jié)果。

思路

思路跟3Sum很像， 不同的就是不用去重復(fù)， 要記錄三和最接近target的和值。

代碼

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int result = 0;
        int minDiff = INT32_MAX;
        // 排序
        sort(nums.begin(), nums.end());
        int n = static_cast<int>(nums.size());
        
        for (int i = 0; i < n - 2; ++i) {
          
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                int diff = abs(sum - target);
                // 記錄三和最接近target的和值
                if (minDiff > diff){
                    result = sum;
                    minDiff = diff;
                }
                else if (sum < target){
                    j++;
                }
                else if (sum > target){
                    k--;
                }
                else{
                    return result;
                }
            }
        }
        return result;
    }
};