zoukankan      html  css  js  c++  java
  • leetcode--3Sum

    Given an array S of n integers, are there elements abc in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

    Note:

    • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
    • The solution set must not contain duplicate triplets.
        For example, given array S = {-1 0 1 2 -1 -4},
    
        A solution set is:
        (-1, 0, 1)
        (-1, -1, 2)

    public class Solution {
        public List<List<Integer>> threeSum(int[] num) {
           List<List<Integer> > result = new ArrayList<List<Integer> >();
    		Arrays.sort(num);
    		int length = num.length;
    		if(length >= 3) {
    			for(int i = 0; i < length - 2 && num[i] <= 0; ++i) {
    				int start = i + 1, end = length - 1;				
    				//find all solution starting with num[i]
    				while(start < end){
    					int sum = num[i] + num[start] + num[end];
    					if(sum < 0) //num[start] is too small
    						++start;
    					else if(sum > 0) //num[end] is too large
    						--end;
    					else { //find a solution
    						List<Integer> oneSolution = new ArrayList<Integer>();
    						oneSolution.add(num[i]);
    						oneSolution.add(num[start]);
    						oneSolution.add(num[end]);
    						result.add(oneSolution);
    						//remove the duplicates 
    						do{
    							++start;
    						}while(start < end && num[start - 1] == num[start]);
    						do{
    							--end;
    						}while(start < end && num[end] == num[end + 1]);
    					}
    				}
    				//remove the duplicates of num[i] since we have found all solutions 
    				//staring with num[i].
    				while(i + 1 < length - 2 && num[i] == num[i + 1])
    					++i;
    			}
    		}
    		return result;
        }
    }
    

      

      

  • 相关阅读:
    Samba
    百度贴吧
    baidu.com关键字查询
    vim 删除每行开头结尾空格
    read line(逐行读取)
    pycharm 激活
    rsync 启动脚本
    收藏
    elk
    sql is null
  • 原文地址:https://www.cnblogs.com/averillzheng/p/3617044.html
Copyright © 2011-2022 走看看