java算法之3Sum

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例题：Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
一、思路最简单的算法就是遍历，时间复杂度会很高，为O（N*N*N）

public class Solution {    public List<List<Integer>> threeSum(int[] nums) {        int len = nums.length;        List<Integer> inlist;        List<List<Integer>> outlist = new ArrayList<List<Integer>>();        for(int i=0;i<len-2;i++){            for(int j=i+1;j<len-1;j++){                for(int k=j+1;k<len;k++){                    if(nums[i]+nums[j]+nums[k]==0){                        inlist = new ArrayList<Integer>();                        inlist.add(nums[i]);                        inlist.add(nums[j]);                        inlist.add(nums[k]);                        Collections.sort(inlist);                        if(!outlist.contains(inlist))                        outlist.add(inlist);                    }                }            }        }       return outlist;    }}

二、降低时间复杂度到O

public class Solution {    public List<List<Integer>> threeSum(int[] nums) {        int len = nums.length;        Arrays.sort(nums);//首先排序        List<List<Integer>> outlist = new ArrayList<List<Integer>>();        for(int i=0;i<len-2;i++){            while(i>0 && nums[i]==nums[i-1] && i<len-2)//跳过相同的元素            i = i+1;            int j= i+1;            int k = len-1;            while(j<k){            if(nums[i]+nums[j]+nums[k] == 0)                 outlist.add(Arrays.asList(nums[i], nums[j], nums[k]));                 j++;                 k--;                 while(nums[j]==nums[j-1] && j<k){ //跳过相同的元素                    j++;                }                 while(nums[k]==nums[k+1] && j<k){ //跳过相同的元素                    k--;                }            }            else if(nums[i]+nums[j]+nums[k] > 0){                k--;            }            else{                j++;            }            }        }       return outlist;    }}

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