LeetCode-3Sum

题目:

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.

思路分析:

• 先将数组按增序排序.
• 遍历数组,设置3个指针i,j,k，先确定i的位置,记b=a[j]=a[i+1],c=a[k]=a[n-1].
• 如果a+b+c>0,则k--,若a+b+c<0，则j++,若a+b+c==0,则保存下来.
• 同时要注意，题目要求not contain duplicate,所以a=a[i]时要判断是否和a[i-1]相等，若相等,则此问题已解决过，同理也要判断b和c相等的情况.

代码:

#include <iostream>
#include <vector>
#include <string>

vector<vector<int> > threeSum(vector<int> &num){
     vector<vector<int> > ret;
     
     if(num.empty()) return ret;
     int size=num.size();
     
     sort(num.begin(),num.end());
     
     for(vector<int>::const_iterator it=num.begin();it!=num.end();it++){
            
            if(it!=nu.begin() && *it==*(it-1)) continue;
            vector<int>::const_iterator front=it+1;
            vector<int>::const_iterator back=num.end()-1;
                
            while(front<back){
                  int sum=*it+*front+*back;
                  
                  if(sum>0){
                      back--;
                  }
                  
                  else if(sum<0){
                      front++;
                  }
                  
                  else if(front!=it+1 && *front==*(front-1)){
                      front++;
                  }
                  
                  else if(back!=num.end() && *back==*(back+1)){
                      back--;
                  }
                  
                  else{
                      vector<int> rsesult;
                      
                      result.push_back(*it);
                      result.push_back(*front);
                      result.push_back(*back);
                      
                      ret.push_back(result);
                      front++;
                      back--;
                  }
            }
     }
     return ret;
}