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).
tag : two pointers
备注：
int closest = Integer.MAX_VALUE / 2; 否则，当全是负数时，容易溢出。
public class Solution {
/* (1) 排序
(2) 把所有三数之和临时存起来，取绝对值最小的，一旦等于0就可以直接返回了。
*/
public int threeSumClosest(int[] nums, int target) {
int closest = Integer.MAX_VALUE / 2;
if(nums == null || nums.length < 3 ) {
return closest;
}
Arrays.sort(nums);
for(int i = 0; i < nums.length - 2; i++) {
int x = nums[i];
int left = i + 1;
int right = nums.length - 1;
while(left < right) {
int y = nums[left];
int z = nums[right];
int sum = x + y + z;
if(sum - target == 0) {
return sum;
}
if(Math.abs(sum - target) < Math.abs(closest-target)) {
closest = sum;
}
if(sum < target) {
left++;
} else {
right--;
}
}
}
return closest;
}
}
原文：http://www.cnblogs.com/superzhaochao/p/6399476.html