题目描述:
Given an array nums
of n integers and an integer target
, find three integers in nums
such that the sum is closest to target
. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
中文理解:
给定一个数和数组,从数组中找到三个元素的和距离给定数最近的数,返回。
解题思路:
首先对数组进行排序,然后外层循环从0到nums.length-2,内层循环则设置两个指针start,end,如果nums[i]+nums[start]+nums[end]大于target则end--,同时比较距离target的距离是否小了,如果nums[i]+nums[start]+nums[end]小于target,则start++,同时比较距离target的距离是否小了,小了更新,如果nums[i]+nums[start]+nums[end]等于target,返回target。
代码(java):
class Solution {
public int threeSumClosest(int[] nums, int target) {
if(nums.length<=3){
int sum=0;
for(int val:nums){
sum+=val;
}
return sum;
}
Arrays.sort(nums);
int res=nums[0]+nums[1]+nums[2];
for(int i=0;i<nums.length-2;i++){
int start=i+1,end=nums.length-1;
while(start<end){
if(target<nums[i]+nums[start]+nums[end]){
if(Math.abs(nums[i]+nums[start]+nums[end]-target)<Math.abs(res-target))res=nums[i]+nums[start]+nums[end];
end--;
}
else if(target>nums[i]+nums[start]+nums[end]){
if(Math.abs(nums[i]+nums[start]+nums[end]-target)<Math.abs(res-target))res=nums[i]+nums[start]+nums[end];
start++;
}
else if(target==nums[i]+nums[start]+nums[end])return target;
}
}
return res;
}
}