[玩转算法]对撞指针

167 Two Sum II - Input array is sorted

给定一个有序整型数组和一个整数target,在其中寻找两个元素，使其和为target。返回这两个数的索引。

-如numbers = [2,7,11,15],target = 9;

-返回数字2,7的索引1，2（索引从1开始计算）

-如果没有解怎么样？保证有解

-如果有多个解怎样？返回任意解

 

#include 
#include 
#include 
#include 
using namespace std;
class Solution{
public:
    vector twoSum(vector& numbers,int target){
        assert(numbers.size() >= 2);
        int l = 0;
        int r = numbers.size() - 1;
        while(l < r){
            if(numbers[l] + numbers[r] == target){
                 int res[] = {l+1,r+1};
                 return vector(res,res+2);
            }
            else if(numbers[l] + numbers[r] < target){
                l++;
            }
            else{
                r--;
            }
        }
        throw ("The input has no solution");
    }
    
};
int main(){
    int arr[] = {2,3,5,7,8,2};
    vector vec(arr,arr+sizeof(arr)/sizeof(int));
    
    vector ret = Solution().twoSum(vec,10);
    for(int i = 0; i < ret.size(); i++){
        cout << ret[i]  << " ";
    }
    cout << endl;
    return 0;
}

 

125.Valid Palindrome

给定一个字符串，只看其中的数字和字母，忽略大小写，判断这个字符串是否为回文串？

·“A man,a plan,a canal;Panama”是回文串

·“race a car”不是回文串

· 空字符串如何看？

· 字符的定义

· 大小写的问题

 

#include 
#include 
#include 
using namespace std;
class Solution{
public:
    bool validPalindrome(vector& vec){
        assert(vec.size() > 0);
        int l = 0;
        int r = vec.size() -2;
        while(l <= r)
        {
            if((vec[l] <= '9' && vec[l] >= '0') || (vec[l] >= 'a' && vec[l] <= 'z') || (vec[l] >= 'A' && vec[l] <= 'Z')){
                if((vec[l] <= '9' && vec[r] >= '0') || (vec[r] >= 'a' && vec[r] <= 'z') ||( vec[r] >= 'A' && vec[r] <= 'Z')){
                    if(vec[l] == vec[r] || vec[l] - 32 == vec[r] || vec[l] + 32 == vec[r]){
                        l++;
                        r--;
                    }
                    else
                    {
                        return false;
                    }
                }
                else{
                    r--;
                }
            }
            else{
                l++;
            }
        }
        return true;
    }
};
int main(){
    //char arr[] = "AmannamA";
    char arr[] = "A man, a plan, a canal; Panama";
    char arr2[] = "a run rice";
    vector vec(arr,arr+sizeof(arr)/sizeof(char));
    vector vec2(arr2,arr2+sizeof(arr2)/sizeof(char));
    bool ret = Solution().validPalindrome(vec);
    bool ret2 = Solution().validPalindrome(vec2);
    if(ret == true){
        cout << "Yes" << endl;
    }
    else{
        cout << "No" << endl;
    }
    
    if(ret2 == true){
        cout << "Yes" << endl;
    }
    else{
        cout << "No" << endl;
    }
    return 0;
}

 

344.Reverse String

给定一个字符串，返回这个字符串的倒序字符串

 

#include 
#include 
#include 
using namespace std;
class Solution{
public:
    void ReverseString(vector& vec){
         int l = 0;
         int r = vec.size() -2;
         while(l <= r){
             swap(vec[l],vec[r]);
             l++;
             r--;
         }
    }
    
};
int main(){
    char arr[] = "helloworld";
    vector vec(arr,arr+sizeof(arr)/sizeof(char));
    Solution().ReverseString(vec);
    for(int i = 0; i < vec.size() - 1; i++){
        cout << vec[i];
    }
    cout << endl;
    return 0;
}

345.Reverse Vowels of a string

给定一个字符串，将该字符串中的元音字母翻转 a e i o u

·如：给出“hello”返回“holle”

·如：给出“leetcode”,返回“leotcede”

·元音字母不包含y

 

#include 
#include 
#include 
using namespace std;
class Solution{
public:
    void ReverseVowels(vector& vec){
         int l = 0;
         int r = vec.size() -2;
         while(l <= r){
             if(vec[l] == 'a' || vec[l] == 'e' || vec[l] == 'i' || vec[l] == 'o' || vec[l] == 'u'){
                 if(vec[r] == 'a' || vec[r] == 'e' || vec[r] == 'i' || vec[r] == 'o' || vec[r] == 'u'){
                     swap(vec[l],vec[r]);
                 }
                 l++;
                 r--;
             }
             else{
                 l++;
                 r--;
             }
         }
    }
    
};
int main(){
    char arr[] = "leetcode";
    vector vec(arr,arr+sizeof(arr)/sizeof(char));
    Solution().ReverseVowels(vec);
    for(int i = 0; i < vec.size() - 1; i++){
        cout << vec[i];
    }
    cout << endl;
    return 0;
}

11.Container With Most Water

给出一个非负整数数组，a1,a2,a3,……，an；每个整数表示一个竖立在坐标轴x位置的一堵高度为ai的“墙”，选择两堵墙，和x轴构成的容器可以容纳最多的水

#include 
#include 
#include 
using namespace std;
class Solution{
public:
    int Min(int a, int b){
        if(a <= b){
            return a;
        }
        else{
            return b;
        }
    }
    vector MosrWater(vector& vec){
        int l = 0;
        int r = vec.size() - 1;
        int result = 0;
        int temp[3];
        vector ret(temp,temp+sizeof(temp)/sizeof(int)) ;
        result =  Min(vec[l],vec[r]) * (r - l);
        ret.clear();
        ret.push_back(result);
        ret.push_back(l);
        ret.push_back(r);
        while(l <= r){
            if((Min(vec[l],vec[r]) * (r - l)) > result){
          	    result =  Min(vec[l],vec[r]) * (r - l);
                ret.clear();
                ret.push_back(result);
                ret.push_back(l);
                ret.push_back(r);
            }
            l++;
            r--;
        }
    	return ret;
    }
};
int main(){
    int arr[] = {4,8,4,7,8,9,10};
    vector vec(arr,arr+sizeof(arr)/sizeof(int));
    vector ret = Solution().MosrWater(vec);
    for(int i = 0; i < ret.size(); i++){
        cout << ret[i] << endl;
    }
    return 0;
}