[LeetCode] 002: 3Sum Closest

最新推荐文章于 2023-02-22 17:17:57 发布

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本文介绍了一个经典的算法问题——寻找数组中三个整数，使其和最接近给定的目标值，并提供了一种高效的解决方案。该算法首先对数组进行排序，然后通过双指针技巧遍历数组以找到最优解。

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[Problem]

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.

    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).

[Analysis]
与 3Sum 类似.

[Solution]

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(num.size() < 3)return 0;
        
        // sort
        sort(num.begin(), num.end());
        
        // search
        int sum = INT_MAX;
        int i = 0, j, k;
        while(i < num.size()){
            // ignore same num[i]
            while(i > 0 && num[i] == num[i-1]){
                i++;
            }
            j = i + 1;
            k = num.size() - 1;
            
            //
            while(j < k){
                // ignore same num[j]
                while(j > i+1 && num[j] == num[j-1] && j < k){
                    j++;
                }
                
                // ignore same num[k]
                while(k < num.size()-1 && num[k] == num[k+1] && j < k){
                    k--;
                }
                if(j >= k)break;
                
                // sum up
                int tmp = num[i] + num[j] + num[k];
                if(tmp == target)return tmp;
                else if(sum == INT_MAX || abs(tmp - target) < abs(sum - target)){
                    sum = tmp;
                }
                
                // next
                if(tmp < target){
                    j++;
                }
                else if(tmp > target){
                    k--;
                }
            }
            i++;
        }
        return sum;
    }
};