LeetCode每日一题：four sum

问题描述

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

问题分析

这题和三个数之和的算法基本一样，没什么变化。

代码实现

public ArrayList> fourSum(int[] num, int target) {
        ArrayList> result = new ArrayList<>();
        if (num.length < 4) return result;
        Arrays.sort(num);
        for (int i = 0; i < num.length - 3; i++) {
            for (int j = i + 1; j < num.length - 2; j++) {
                int left = j + 1;
                int right = num.length - 1;
                while (left < right) {
                    int sum = num[left] + num[right];
                    if (sum < target - num[i] - num[j]) left++;
                    if (sum > target - num[i] - num[j]) right--;
                    if (sum == (target - num[i] - num[j])) {
                        ArrayList list = new ArrayList<>();
                        list.add(num[i]);
                        list.add(num[j]);
                        list.add(num[left]);
                        list.add(num[right]);
                        result.add(list);
                        int temp1 = num[left];
                        int temp2 = num[right];
                        while (left < right && num[left] == temp1) left++;
                        while (left < right && num[right] == temp2) right--;
                    }
                }
                while (j + 1 < num.length - 2 && num[j + 1] == num[j]) j++;
            }
            while (i + 1 < num.length - 3 && num[i + 1] == num[i]) i++;
        }
        return result;
    }