【LeetCode】3Sum Closest 解题报告
2017-06-11 12:20
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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.
【解析】
和
3Sum解题报告 非常像,与之不同的是,不再是求三个数的和是不是为0,而是看三个数的和与target的差是否为最小,仅仅需记录当前最优解并不断更新其值就可。
【Java代码】O(n^2)
easy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),这样会导致溢出!。
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).
【解析】
和
3Sum解题报告 非常像,与之不同的是,不再是求三个数的和是不是为0,而是看三个数的和与target的差是否为最小,仅仅需记录当前最优解并不断更新其值就可。
【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 < len-2; i++) { if (i > 0 && num[i] == num[i-1]) continue; int l = i+1, r = len-1; 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; } }
easy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),这样会导致溢出!。
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