标签：i++ logs pre == ret cti continue hat any
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).
这道题目应该是3Sum的一个简化吧，主要也是使用两个指针向中间遍历的方法。代码如下：
public class Solution {
public int threeSumClosest(int[] nums, int target) {
int len = nums.length;
int ans = 0;
boolean bans = false;
Arrays.sort(nums);
for(int i=0;i<len-1;i++){
if(i>0 && nums[i] == nums[i-1])
continue;
int left = i+1,right = len-1;
while(left<right){
int sum = nums[i]+nums[left]+nums[right];
if(sum == target){
return ans = sum;
}
else if( sum > target ){
right--;
}
else{
left++;
}
if(!bans){
ans = sum;
bans = true;
}
else if( Math.abs(target-ans) > Math.abs(target-sum) ){
ans = sum;
}
}
}
return ans;
}
}
Leetcode Array 16 3Sum Closest
标签：i++ logs pre == ret cti continue hat any
原文地址：http://www.cnblogs.com/mxk-star/p/7252708.html
