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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).
此题与上体3sum类似，只不过这里需要用sum来和target比较决定指针+1,还是-1,同样这里需要先对元素排序，代码如下：
void insertsort2(vector<int> &num) { int nSize = num.size(); int j = 0; for(int i = 1; i< nSize; ++i) { int temp = num[i]; for( j = i; j>0 && temp < num[j-1] ; --j) { num[j] = num[j-1]; } num[j] = temp; } } int threeSumClosest(vector<int>& nums, int target) { int nSize = nums.size(); int sum = 0; int dist = INT_MAX; int nPrevDist = INT_MAX; insertsort2(nums); if(nSize < 3) { return -1; } for(int i = 0; i!=nSize; ++i) { int p = i+1; int q = nSize - 1; while(p < q) { sum = nums[i] + nums[p] + nums[q]; int temp = abs(sum - target); if(temp < nPrevDist) { nPrevDist = temp; dist = sum; } if(sum > target) { --q; } else { ++p; } } } return dist; }
需要多思考如何借鉴运用已有的算法。
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原文地址：http://www.cnblogs.com/bestwangjie/p/4458518.html
