3Sum Closet

3Sum ClosestJan 18 '12

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).

算法：可以采用和3Sum问题相似的算法，对数组排序后枚举v[i]，取1-v[i]作为target，进行类似2Sum的计算，返回与target差值最小的差值。算法复杂度也是O(N^2)的。

程序：

int threeSumClosest(vector<int> &num, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        sort(num.begin(),num.end());            int target2sum;int minum_difference;int flag=0;vector<int>::iterator it = num.begin();for(;it!=num.end()-1;){target2sum = target-*it;;if(it==num.begin()) flag = 1;//erase the elementit = num.erase(it);//2Sum2if(flag){minum_difference = twoSumCloset(num,target2sum);flag = 0;}else{int ret2 = twoSumCloset(num,target2sum);minum_difference = abs(minum_difference)<abs(ret2)?minum_difference:ret2;}num.insert(it,target-target2sum);it++;}return minum_difference+target;    }        int twoSumCloset(vector<int> &num,int target){        int i = 0;        int j = num.size()-1;int difference = num[i]+num[j]-target;        int ret = difference;        while(i<j){            difference = num[i]+num[j]-target;if(abs(difference)<abs(ret)) ret = difference;            if(difference > 0){                j--;            }else if(difference<0){                i++;            }else{                break;            }        }        return ret;    }

 时间：112 milli secs