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]
]

解析

问题一：

为何第一个for loop的查重 if 条件是 i>0 ， 第二个for loop的查重 if 条件是 j>i+1因为要查重，至少保证有2个元素，比如：i = 0 和 i >0 开始查重。

思路

1. two pointers method
2. the first pointer named i , it start from 0 to nums.length - 3, sweeping the whole array
3. meanwhile, the second pointer named j , it start from i+1 to nums.length - 2, sweeping the whole array
4. For each possible first element we make a standard bi-directional 2Sum sweep of the remaining part of the array
5. At the same time, we need to skip equal elements to avoid duplicates

图解

代码

class Solution {
       public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums.length < 4) return result;
        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 k = j + 1;
                int l = nums.length - 1;
                while (k < l) {
                    final int sum = nums[i] + nums[j] + nums[k] + nums[l];
                    if (sum < target) {
                        ++k;
                        while(nums[k] == nums[k-1] && k < l) ++k;
                    } else if (sum > target) {
                        --l;
                        while(nums[l] == nums[l+1] && k < l) --l;
                    } else {
                        result.add(Arrays.asList(nums[i], nums[j], nums[k], nums[l]));
                        ++k;
                        --l;
                        while(nums[k] == nums[k-1] && k < l) ++k;
                        while(nums[l] == nums[l+1] && k < l) --l;
                    }
                }
            }
        }
        return result;
    }
}