【数据结构与算法】十三 twoSum hash

【数据结构与算法】十三 twoSum hash

无序数组两数和等于目标

C++

#include <stdio.h> #include <iostream> using namespace std ; template <class T> int getArrayLen(const T& array){ return sizeof(array) / sizeof(array[0]) ; } template <class T> void twoSum(int target , T & array){ int len = getArrayLen(array) ; for(int i=0;i<len;i++){ for( int j=i+1;j<len;j++){ if(array[i] + array[j] == target){ cout << "found " << array[i] << " , " << array[j] << endl; } } } cout << endl; } int main(){ int target = 10 ; int a1[] = {0,1,2,3,4,5,6,7,8,9}; twoSum(target , a1); int a2[] = {5,6,7,8,9,0,1,2,3,4}; twoSum(target , a2); return 0; } 

以上是双重循环，很容易想到，但是效率问题，下面来优化

时间复杂度

O(n2)

排序后求和

C++

#include <stdio.h> #include <iostream> #include <algorithm> using namespace std ; template <class T> int getArrayLen(const T& array){ return sizeof(array) / sizeof(array[0]) ; } template <class T> void twoSum(int target , T & array){ int len = getArrayLen(array) ; int left = 0 ; int right = len - 1; while( left < right ){ if(array[left] + array[right] == target){ cout << "found " << array[left] << " , " << array[right] << endl; left ++ ; right -- ; }else if(array[left] + array[right] < target) left ++; else right --; } cout << endl; } int main(){ int target = 9 ; int a1[] = {0,1,7,8,9,2,3,4,5,6}; sort(a1, a1 + 10); twoSum(target , a1); int a2[] = {5,6,7,8,9,0,1,2,3,4}; sort(a2, a2 + 10); twoSum(target , a2); return 0; } 

时间复杂度

O(nlogn)

基于hash实现

C++

#include <stdio.h> #include "iostream" #include <unordered_set> using namespace std; template <class T> int getArrayLen(const T& array){ return sizeof(array) / sizeof(array[0]) ; } template <class T> void twoSum(int target , T & array){ unordered_set<int> hash_set; int len = getArrayLen(array); for(int i = 0; i < len; i++){ if(hash_set.count(target - array[i])){ cout << "Found:" << array[i] << "," << target-array[i] << endl; } hash_set.insert(array[i]); } cout << endl; } int main(){ int target = 9; int a1[] = {0,1,7,8,9,2,3,4,5,6}; int a2[] = {5,6,7,8,9,0,1,2,3,4}; twoSum(target , a1); twoSum(target , a2); return 0; } 

时间复杂度

O(n)

JAVA

    static void twoSum(int target , int array[]){
        int len = array.length;
        HashSet set = new HashSet();
        for(int i=0;i<len;i++){
            if(set.contains( target - array[i] ) )
                System.out.println("fond : " + array[i] + " , " + (target - array[i]) );
            set.add(array[i]);
        }
    }

    public static void main(String[] args) {
        int target = 9;
        int a1[] = {3,7,0,1,2,8,9,4,5,6};
        twoSum(target , a1);
        System.out.println();
        int a2[] = {1,2,8,9,4,5,6,3,7,0};
        twoSum(target , a2);
    }

时间复杂度

O(n)