Leetcode - 3Sum Closest

最新推荐文章于 2024-04-29 23:48:08 发布

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本文介绍了一种寻找数组中三个整数使它们的和最接近指定目标值的有效算法。通过排序和双指针技巧，实现了O(n^2)的时间复杂度。

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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;
    }