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LeetCode：3Sum Closest
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).
Solution：JAVA
public class Solution { public int threeSumClosest(int[] num, int target) { int min = Integer.MAX_VALUE; int result = 0; Arrays.sort(num); for (int i = 0; i < num.length; i++) { int j = i + 1; int k = num.length - 1; while (j < k) { int sum = num[i] + num[j] + num[k]; int diff = Math.abs(sum - target); if (diff < min) { min = diff; result = sum; } if (sum <= target) { j++; } else { k--; } } } return result; }}
C++
class Solution {public: int threeSumClosest(vector<int> &num, int target) { // Note: The Solution object is instantiated only once and is reused by each test case. if(num.empty()) return 0; sort(num.begin(), num.end()); int min= INT_MAX; int record; for(int i=0; i<num.size(); i++) { int tmp = target - num[i]; int start = i+1, end = num.size()-1; while(start<end) { int sum = num[start]+num[end]+num[i]; if(sum==target) { min = 0; record = sum; break; } else if(sum > target) { if(abs(target-sum)<min) { min = abs(target-sum); record = sum; } end--; } else if(sum < target) { if(abs(target-sum)<min) { min = abs(target-sum); record = sum; } start++; } while(i<num.size()-1&&num[i]==num[i+1]) i++; } } return record； }};
对数组求最大值，联想预处理排序
LeetCode：3Sum Closest
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