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3Sum Closest

2024-07-17 20:16:15   213人阅读

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).

分析：跟3sum类似的方法，使用三个指针，让最左边的指针l从[0,size-2]变化，对于每一个固定的最左指针l，中间指针和最右指针在[l+1,size-1]范围内移动，移动方法跟2sum一致。先初始化min_dif=INT_MAX，如果出现更小的dif，则更新min_dif以及closest sum。时间复杂度是O(n^2)，空间复杂度是O(1)。
代码如下：
 1 class Solution { 2 public: 3     int threeSumClosest(vector<int> &num, int target) { 4         int size = num.size(); 5         if(size < 3) return 0; 6         sort(num.begin(),num.end()); 7         int min_dif = INT_MAX; 8         int closest_sum; 9         int l = 0, mid = 1, r = size-1;10         for(;l < size-2;l++){11             mid = l + 1;12             r = size - 1;13             while(mid < r){14             int sum = num[l] + num[mid] + num[r];15             int dif = abs(sum-target);16             if(dif < min_dif){17                 closest_sum = sum;18                 min_dif = dif;19             }20             if(sum < target) mid++;21             else r--;22             }23         }24         return closest_sum;25     }26 };

 


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