18. 4Sum

最新推荐文章于 2021-04-19 21:52:06 发布

sunnylinner

最新推荐文章于 2021-04-19 21:52:06 发布

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分类专栏： LeetCode 文章标签： java LeetCode string

本文链接：https://blog.youkuaiyun.com/sunnylinner/article/details/51144914

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本文详细阐述了解决四数之和问题的算法，即在一个整数数组中查找四个元素的组合，使得它们的和等于特定的目标值。文章包括排序数组、递归搜索和避免重复解决方案的方法，确保输出的四元组按升序排列且不包含重复项。

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Given an array S of n integers, are there elements a,b,c, andd 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)

public class Solution {
	public List<List<Integer>> fourSum(int[] nums, int target) {
		List<List<Integer>> result = new ArrayList<List<Integer>>();
		Arrays.sort(nums);
		kSum(result, new ArrayList<Integer>(), nums, target, 4, 0, nums.length-1);
		return result;
	}
	
	private void kSum(List<List<Integer>> result,List<Integer> cur, int[] nums, int target,
					int k, int start, int end) {
		if(k == 0 || nums.length ==0 || start>end)
			return;
		if(k == 1){
			for (int i = start; i <= end; i++) {
				if (nums[i] == target) {
					cur.add(nums[i]);//添加
					result.add(new ArrayList<Integer>(cur));//放入到结果中
					cur.remove(cur.size()-1);//移除
				}
			}
			return;
		}
		
		if(k == 2){
			int sum;
			while (start < end){
				sum = nums[start] + nums[end];
				
				if(sum == target){
					cur.add(nums[start]);
					cur.add(nums[end]);
					result.add(new ArrayList<Integer>(cur));
					cur.remove(cur.size()-1);
					cur.remove(cur.size()-1);
					
					//避免重复
					while(start < end && nums[start] == nums[start+1])
						++start;
					++start;
					while(start < end && nums[end] == nums[end-1])
						--end;
					--end;
				}else if(sum < target){
					//避免重复
					while(start < end && nums[start] == nums[start+1])
						++start;
					++start;
				}else{
					while(start < end && nums[end] == nums[end-1])
						--end;
					--end;
				}
			}
			return;
		}
		
		//每次务必会进行数据预处理，简单的情况判断，避免了不必要的情况
		if(k*nums[start] > target || k*nums[end] < target) 
			return;
		
		for(int i = start; i <= (end-k+1); i++){
			if(i > start && nums[i] == nums[i-1])
				continue;//避免了重复
			if(k*nums[i] <= target) {
				cur.add(nums[i]);
				kSum(result, cur, nums, target - nums[i], k - 1, i + 1, end);
				//开始了神奇的递归了
				cur.remove(cur.size() - 1);//输入的地方总要对应一个输出
			}
		}
		
	}
}
//k=0,1太简单了,k=2,见上一题

这题跟之前的几道题是差不多的~