Java [leetcode 18]4Sum

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, abcd)
  • 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)

解题思路:

与之前的3sum相类似,只不过在外边多了一套循环,时间复杂度为O(n^3)

代码如下:

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public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    public List<List<Integer>> fourSum(int[] nums, int target) {
        int length = nums.length;
        int tmp;
        if (nums == null || length < 4)
            return ans;
        Arrays.sort(nums);
        for (int i = 0; i < length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            for (int j = i + 1; j < length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1])
                    continue;
                int begin = j + 1;
                int end = length - 1;
                while (begin < end) {
                    tmp = nums[begin] + nums[end] + nums[i] + nums[j];
                    if (tmp == target) {
                        List<Integer> list = new ArrayList<Integer>();
                        list.add(nums[i]);
                        list.add(nums[j]);
                        list.add(nums[begin]);
                        list.add(nums[end]);
                        ans.add(list);
                        while (begin < end && nums[begin + 1] == nums[begin])
                            begin++;
                        begin++;
                        while (begin < end && nums[end - 1] == nums[end])
                            end--;
                        end--;
                    } else if (tmp > target)
                        end--;
                    else
                        begin++;
                }
            }
        }
        return ans;
    }
}

 

posted @   scottwang  阅读(262)  评论(0编辑  收藏  举报
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