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[LeetCode] 3Sum Closest

2024-07-17 00:58:06   220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：和3Sum一样，不能用O（n^3）遍历，需要有一些可以简化的步骤，三层遍历肯定Time Limited Exceeded！
class Solution {public:    int threeSumClosest(vector<int> &num, int target) {        int n = num.size();        map<int,int> diff;//key是target和sum的差值的绝对值，value是sum值        sort(num.begin(),num.end());        int sum ;        for(int i=1;i<n-1;i++){               int l = 0, r= n-1;               while(l<i && r>i){               if(l>1 && num[l] == num[l-1]){                  l++;                  continue;               }               if(r<n-1 && num[r] == num[r+1]){                  r--;                  continue;               }               sum = num[l]+num[i]+num[r];               if(sum == target)                   return sum;               else {                   int abs   = sum-target>0 ? sum-target :target-sum;                   diff[abs] = sum;                   if(sum >target)                       r--;                   else                       l++;                                      }           }//end while        }//end for        map<int,int>::iterator iter = diff.begin();        return (*iter).second;    }};

 


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