【LeetCode】4Sum 解题报告

【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:

• 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)

【解析】

3Sum 和 3Sum Closest 的扩展，相同思路，加强理解。

K Sum 问题的时间复杂度好像为 O(n^(k-1)) ？！假设有更好的，欢迎不吝赐教！

【Java代码】

public class Solution {
    List<List<Integer>> ret = new ArrayList<List<Integer>>();
    
    public List<List<Integer>> fourSum(int[] num, int target) {
        if (num == null || num.length < 4) return ret;
        Arrays.sort(num);
        int len = num.length;
        for (int i = 0; i < len-3; i++) {
            if (i > 0 && num[i] == num[i-1]) continue;
            for (int j = i+1; j < len-2; j++) {
                if (j > i+1 && num[j] == num[j-1]) continue;
                findTwo(num, j+1, len-1, target, num[i], num[j]);
            }
        }
        return ret;
    }
    
    public void findTwo(int[] num, int begin, int end, int target, int a, int b) {
        if (begin < 0 || end >= num.length) return;
        int l = begin, r = end;
        while (l < r) {
            if (a+b+num[l]+num[r] < target) {
                l++;
            } else if (a+b+num[l]+num[r] > target) {
                r--;
            } else {
                List<Integer> ans = new ArrayList<Integer>();
                ans.add(a);
                ans.add(b);
                ans.add(num[l]);
                ans.add(num[r]);
                ret.add(ans);
                l++;
                r--;
                while (l < r && num[l] == num[l-1]) l++;
                while (l < r && num[r] == num[r+1]) r--;
            }
        }
    }
}