Leetcode - 3Sum Closest

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).

[分析] 基于 3 sum的思路求解,先排序,选定第一个元素,然后夹逼法求出当前第一个元素下最接近target的 sum3。注意到closest初始化不要写成整数的最大最小值,不然比较获取局部最优时(if (Math.abs(sum3 - target) < Math.abs(closest - target)))会溢出导致错误结果。

public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3)
            return Integer.MAX_VALUE;
        int N = nums.length;
        Arrays.sort(nums);
        int closest = nums[0] + nums[1] + nums[2];
        int sum3;
        for (int k = 0; k < N - 2; k++) {
            if (k > 0 && nums[k] == nums[k - 1])
                continue;
            int i = k + 1, j = N - 1;
            while (i < j) {
                sum3 = nums[i] + nums[j] + nums[k];
                if (sum3 == target) {
                    return target;
                } else if (sum3 > target) {
                    if (Math.abs(sum3 - target) < Math.abs(closest - target))
                        closest = sum3;
                    while (i < --j && nums[j] == nums[j + 1]);
                } else {
                    if (Math.abs(sum3 - target) < Math.abs(closest - target))
                        closest = sum3;
                    while (++i < j && nums[i] == nums[i - 1]);
                }
            }
        }
        return closest;
    }

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转载自likesky3.iteye.com/blog/2228875