- 思路
- 2sum的基础上外加一层循环，用于遍历第1~n-2的n-2个数
- 首先对num数组进行排序
- 从第二个数j和最后一个数k开始向中间夹逼
- 如果遇到两个相等的数向下一个移动
- 维护一个minus数用于表示sum和target的差值
- 如果等于0，那么返回target
- 如果小于0，j++
- 如果大于0，k--
- 另外维护一个dis=minus的绝对值
- 如果每一次遍历使得dis变小，那么将新的sum赋值给ret
package leetcode.doit;
import java.util.Arrays;
/**
* 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).
*
*/
public class ThreeSumClosest {
public int threeSumClosest(int[] num, int target) {
int dis = Integer.MAX_VALUE;
int ret = 0;
Arrays.sort(num);
int length = num.length;
for (int i = 0; i < length - 2; i++) {
int j = i + 1;
int k = length - 1;
while (j < k) {
if (j > i + 1 && num[j] == num[j - 1]) {
j++;
continue;
}
if (k < length - 1 && num[k] == num[k + 1]) {
k--;
continue;
}
int sum = num[i] + num[j] + num[k];
int minus = sum - target;
int d = Math.abs(minus);
if (d < dis) {
dis = d;
ret = sum;
}
if (minus == 0)
return target;
if (minus < 0) {
j++;
} else {
k--;
}
}
}
return ret;
}
}