LeetCode 016 3Sum Closest
2014-02-13 22:46
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题目
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.
给一组数和一个目标数,找到其中3个数,这3个数满足条件相加与目标数最接近。输出这样的三个数之和。
思路
和之前的3sum方法一样 http://blog.csdn.net/xift810/article/details/19175113
有些细节要变化一下:
a 需要有个差值记录器,每当差距缩小的时候,都要记录下来。
b 当3个数总和相同时候,直接输出答案。
c 最开始的差值用系统最大值。
代码
15.3.1.8 注意 if(sum==target){
return target;
}
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).
给一组数和一个目标数,找到其中3个数,这3个数满足条件相加与目标数最接近。输出这样的三个数之和。
思路
和之前的3sum方法一样 http://blog.csdn.net/xift810/article/details/19175113
有些细节要变化一下:
a 需要有个差值记录器,每当差距缩小的时候,都要记录下来。
b 当3个数总和相同时候,直接输出答案。
c 最开始的差值用系统最大值。
代码
public int threeSumClosest(int[] num, int target) { if(num==null|| num.length<3){ return 0; } Arrays.sort(num); int len = num.length; int leaf = Integer.MAX_VALUE; int ans=target; for(int i=0;i<len-2;i++){ if(i>0&&num[i]==num[i-1]){ continue; } int l = i +1; int r = len-1; while(l<r){ int sum= num[i] + num[l] + num[r]; if (Math.abs(sum-target)<leaf) { leaf=Math.abs(sum-target); ans = sum; } if(sum==target){ return target; } else if(sum < target){ do{ l++; }while(l<r&&num[l-1]==num[l]); } else{ do{ r--; }while(l<r&&num[r]==num[r+1]); } } } return ans; }
15.3.1.8 注意 if(sum==target){
return target;
}
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