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:

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 @ 2015-05-10 16:40  scottwang  阅读(262)  评论(0编辑  收藏  举报