【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.
For example, given array S = {⑴ 2 1 ⑷}, and target = 1.
The sum that is closest to the target is 2. (⑴ + 2 + 1 = 2).
【解析】
和
3Sum解题报告 很像，与之不同的是，不再是求3个数的和是否是为0，而是看3个数的和与target的差是不是为最小，只需记录当前最优解其实不断更新其值便可。
【Java代码】O(n^2)
public class Solution {
public int threeSumClosest(int[] num, int target) {
if (num == null || num.length < 3) return 0;
Arrays.sort(num);
int ret = 0;
int closestDist = Integer.MAX_VALUE;
int len = num.length;
for (int i = 0; i < len⑵; i++) {
if (i > 0 && num[i] == num[i⑴]) continue;
int l = i+1, r = len⑴;
while (l < r) {
int sum = num[i] + num[l] + num[r];
if (sum < target) {
if (target-sum < closestDist) {
closestDist = target - sum;
ret = sum;
}
l++;
} else if (sum > target) {
if (sum-target < closestDist) {
closestDist = sum - target;
ret = sum;
}
r--;
} else { //when sum == target, return sum.
return sum;
}
}
}
return ret;
}
}
容易出错的地方是，把 ret 初始值设为 Integer.MAX_VALUE，然后后面计算 closestDist = Math.abs(ret - target)，这样会致使溢出！！