3Sum

标签：3sum

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.

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) {
        int len = num.length;
        List<List<Integer>> list = new LinkedList<List<Integer>>();
        if(len<3){
            return list;
        }
        Arrays.sort(num);
        int i=0;
        while(i<len-2){
            int s=i+1;
            int e=len-1;
            while(s<e){
                if(num[s]+num[e]==0-num[i]){
                    List<Integer> lst = new LinkedList<Integer>();
                    lst.add(num[i]);
                    lst.add(num[s]);
                    lst.add(num[e]);
                    list.add(lst);
                    while(s+1<len&&num[s+1]==num[s]){
                        s++;
                    }
                    s++;
                    while(e-1>=0&&num[e-1]==num[e]){
                        e--;
                    }
                    e--;
                }else if(num[s]+num[e]<0-num[i]){
                    s++;
                }else{
                    e--;
                }
            }
            while(i+1<len&&num[i+1]==num[i]){
                i++;
            }
            i++;
        }
        return list;
    }
}

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > r;
        int n=num.size();
        sort(num.begin(),num.end());
        if(n<3)
            return r;
        
        int i=0;
        while(i<n-2){
            int k1 = num[i];
            int s = i+1,e=n-1;
            while(s<e){
                if(num[s]+num[e]==0-k1){
                    vector<int> v;
                    v.push_back(k1);
                    v.push_back(num[s]);
                    v.push_back(num[e]);
                    
                    r.push_back(v);
                    while(s+1<=n-1&&num[s]==num[s+1]){
                        s++;
                    }
                    s++;
                    while(e-1>=i&&num[e]==num[e-1]){
                        e--;
                    }
                    e--;
                }else if(num[s]+num[e]<0-k1){
                    s++;
                }else{
                    e--;
                }
            }
            while(i+1<=n-1&&num[i]==num[i+1]){
                i++;
            }
            i++;
        }
        return r;
    }
};

3Sum

标签：3sum

原文地址：http://blog.csdn.net/u010786672/article/details/41148131