zoukankan      html  css  js  c++  java
  • [leetcode] 4Sum

    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)

    https://oj.leetcode.com/problems/4sum/

    思路:转换成2Sum,注意跟3Sum一样先排序去重。

    import java.util.ArrayList;
    import java.util.Arrays;
    
    public class Solution {
    	public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
    		ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
    		if (num == null || num.length < 4)
    			return result;
    
    		int n = num.length;
    		Arrays.sort(num);
    		int i, j, start, end, sum;
    		for (i = 0; i < n - 3; i++) {
    			for (j = i + 1; j < n - 2; j++) {
    				start = j + 1;
    				end = n - 1;
    				while (start < end) {
    					sum = num[start] + num[end];
    					if (sum < target - num[i] - num[j])
    						start++;
    					else if (sum > target - num[i] - num[j])
    						end--;
    					else {
    						ArrayList<Integer> tmp = new ArrayList<Integer>();
    						tmp.add(num[i]);
    						tmp.add(num[j]);
    						tmp.add(num[start]);
    						tmp.add(num[end]);
    						result.add(tmp);
    						start++;
    						end--;
    						while (start < j && num[start - 1] == num[start])
    							start++;
    						while (end >= j + 1 && num[end + 1] == num[end])
    							end--;
    
    					}
    				}
    				while (j < n - 1 && num[j] == num[j + 1])
    					j++;
    			}
    			while (i < n - 1 && num[i] == num[i + 1])
    				i++;
    
    		}
    		return result;
    	}
    
    	public static void main(String[] args) {
    		System.out.println(new Solution().fourSum(new int[] { 1, 0, -1, 0, -2,
    				2 }, 0));
    
    	}
    }

    第二遍记录:

      别忘记先排序。

      去重方法类似3Sum。

    public class Solution {
        public List<List<Integer>> fourSum(int[] num, int target) {
            List<List<Integer>> res = new ArrayList<List<Integer>>();
            if(num==null||num.length<4)
                return res;
            int n = num.length;
            Arrays.sort(num);
            
            for(int i=0;i<n-3;i++){
                for(int j=i+1;j<n-2;j++){
                    int start = j+1;
                    int end = n-1;
                    
                    while(start<end){
                        if(num[start]+num[end]<target-num[i]-num[j]){
                            start++;
                        }else if(num[start]+num[end]>target-num[i]-num[j]){
                            end--;
                        }else{
                            List<Integer> tmp = new ArrayList<Integer>();
                            tmp.add(num[i]);
                            tmp.add(num[j]);
                            tmp.add(num[start]);
                            tmp.add(num[end]);
                            res.add(tmp);
                            start++;
                            end--;
                            while(start<n&&num[start]==num[start-1])
                                start++;
                            while(end>=0&&num[end]==num[end+1])
                                end--;
                            
                        }
    
                    }
                    while(j<n-1&&num[j+1]==num[j])
                        j++;
                    
                }
                while(i<n-1&&num[i]==num[i+1])
                    i++;
                
            }
            return res;
        }
    }
  • 相关阅读:
    DHCP DHCPv6
    DHCPv6协议
    IPv6邻居发现协议
    CentOS下禁止防火墙
    centOS下更新yum源
    centOS下yum报错
    Flink+Kafka整合的实例
    Flink基本概念
    Ubuntu16.04下配置ssh免密登录
    Zookeeper+Kafka的单节点配置
  • 原文地址:https://www.cnblogs.com/jdflyfly/p/3810696.html
Copyright © 2011-2022 走看看