• leet_15


    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)
    package com.mingxin.leetcode.leet_15;
    
    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.List;
    
    /**
     * Created by Administrator on 2016/1/25.
     */
    public class ThreeSum {
        public static void main(String[] args){
            int num[] = {-1, -20, -5, -7, 0, 1, 2, 7, 6, 10};
            List<List<Integer>> result = threeSum(num);
            for(List<Integer> item1:result){
                System.out.println("{" + item1.get(0) + ", " + item1.get(1) + ", " + item1.get(2) + "}");
            }
        }
    
        public static List<List<Integer>> threeSum(int[] nums) {
            List<List<Integer>> res = new ArrayList<List<Integer>>();
            if(nums.length<3||nums == null){
                return new ArrayList<List<Integer>>();
            }
    
            Arrays.sort(nums);
            for(int i = 0; i <= nums.length-3; i++){
                if(i==0||nums[i]!=nums[i-1]){
                    int low = i+1;
                    int high = nums.length-1;
                    while(low<high){
                        int sum = nums[i]+nums[low]+nums[high];
                        if(sum == 0){
                            ArrayList<Integer> unit = new ArrayList<Integer>();
                            unit.add(nums[i]);
                            unit.add(nums[low]);
                            unit.add(nums[high]);
    
                            res.add(unit);
    
                            low++;
                            high--;
    
                            while(low<high&&nums[low]==nums[low-1]){
                                low++;
                            }
    
                            while(low<high&&nums[high]==nums[high+1]){
                                high--;
                            }
                        }else if(sum > 0){
                            high --;
                        }
                        else{
                            low ++;
                        }
    
                    }
                }
            }
            return res;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/kniught-ice/p/5159581.html
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