18. 4Sum

题目：

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: 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]
]

链接：https://leetcode.com/problems/4sum/#/description

4/11/2017

88ms, 28%

写错了无数遍。

注意的问题：

1. 外层的遍历并不是从头尾一起开始的，而是都是从一个方向开始的。注意第7行的终止条件

2. while循环当中的判断，并且注意28, 29行

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        if (nums == null || nums.length < 4) return new ArrayList<>();
        List<List<Integer>> ret = new ArrayList<>();

        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            for (int j = i + 1; j < nums.length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                int left = j + 1, right = nums.length - 1;

                while (left < right) {
                    int sum = nums[left] + nums[right] + nums[i] + nums[j];
                    if (sum == target) {
                        List<Integer> l = new ArrayList<Integer>();
                        l.add(nums[i]);
                        l.add(nums[left]);
                        l.add(nums[right]);
                        l.add(nums[j]);
                        ret.add(l);
                        left++;
                        right--;
                        while (left < right && nums[left] == nums[left - 1]) left++;
                        while (left < right && nums[right] == nums[right + 1]) right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return ret;
    }
}

别人的做法：

把kSum变为2sum来解。

https://discuss.leetcode.com/topic/29585/7ms-java-code-win-over-100/17

还有解释时间复杂度lower case

https://discuss.leetcode.com/topic/27445/lower-bound-n-3

更多讨论：

https://discuss.leetcode.com/category/26/4sum