3Sum Closet

最新推荐文章于 2021-07-02 14:12:25 发布

zephyr_be_brave

最新推荐文章于 2021-07-02 14:12:25 发布

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分类专栏： leetcode习题分析

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本文介绍了一个算法问题——寻找数组中三个整数的组合，使其和最接近给定的目标值，并提供了一种解决方案。通过对数组排序并使用类似两数之和的方法来枚举，最终找到与目标值差值最小的三数之和。

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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 element
			it = num.erase(it);
			//2Sum2
			
			if(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