相關問題總結：

• 二分查詢程式中，思想簡單，但是要注意邊界條件。

#include <iostream>
#include <vector>
using namespace std;
// 非遞迴版本
int Binary_search1(vector<int> num, int target)
{
    int size = num.size();
    if (size <= 0)
        return -1;
    int left = 0;
    int right = size;
    int mid;
    while (left < right)
    {
        mid = left + ((right - left) >> 1
 
);
        if (num[mid] > target)
            right = mid;
        else if (num[mid] < target)
            left = mid + 1;
        else
            return mid;
    }
    return -1;
}
// 非遞迴版本
int Binary_Search(vector<int> num, int target)
{
    int size = num.size();
    if (size <= 0)
        return
 
 -1;
    int left = 0;
    int right = size - 1;  // 此處控制迴圈結束的條件
    int mid;
    while (left <= right)  // 依據right的值的變化來確定的
    {
        mid = left + ((right - left) >> 1);   // 防止溢位，同時移位更高效。注意每次迴圈都需要更新
        if (num[mid] > target)
            right = mid - 1;   // right的賦值，隨著right的初值進行改變。
        else
 
 if (num[mid] < target)
            left = mid + 1;
        else
            return mid;
    }
    return -1;
}
int main()
{
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(3);
    nums.push_back(5);
    nums.push_back(5);
    nums.push_back(5);
    nums.push_back(7);
    nums.push_back(10);
    int target = 5;
    int result = Binary_Search(nums, target);
    cout << result << endl;
}

• 最長遞增子序列問題，利用動態規劃實現

#include <iostream>
#include <vector>
using namespace std;
int findLongest(vector<int> num, int n)
{
    if (n <= 0)
        return 0;
    vector<int> dp(n, 0);
    int result = INT_MIN;
    for (int i = 0; i < n; i++)
    {
        dp[i] = 1;
        for (int j = 0; j < i; j++)
        {
            if (num[i] >= num[j])
                dp[i] = max(dp[i], dp[j] + 1);
        }
        if (dp[i] > result)
                result = dp[i];
    }
    return result;
}
int main()
{
    vector<int> num;
    num.push_back(2);
    num.push_back(1);
    num.push_back(4);
    num.push_back(3);
    num.push_back(1);
    num.push_back(5);
    num.push_back(6);
    int n = 7;
    int result = findLongest(num, n);
    cout << result << endl;
}

• 最長公共子序列問題，利用dp實現

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int findLCS(string A, int n, string B, int m)
{
    vector< vector<int> > dp(n, vector<int> (m, 0));
    if (A[0] == B[0])
        dp[0][0] = 1;
    else
        dp[0][0] = 0;
    // 初始化第一列
    for (int i = 1; i < n; i++)
        dp[i][0] = max(dp[i - 1][0], A[i] == B[0] ? 1 : 0);
    // 初始化第一行
    for (int j = 1; j < m; j++)
        dp[0][j] = max(dp[0][j - 1], A[0] == B[j] ? 1 : 0);
    for (int i = 1; i < n; i++)
    {
        for (int j = 1; j < m; j++)
        {
            dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            if (A[i] == B[j])
                dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
        }
    }
    return dp[n - 1][m - 1];
}
int main()
{
    string A = "1A2C3D4B56";
    string B = "B1D23CA45B6A";
    int n = 10;
    int m = 12;
    int result = findLCS(A, n, B, m);
    cout << result << endl;
}

• 最長公共子串。利用動態規劃求解

#include <iostream>
#include <string>
#include <vector>
using namespace std;
int findLongest(string A, int n, string B, int m)
{
    vector< vector<int> > dp(n, vector<int>(m, 0));
    int max_result = 0;
    // 初始化dp
    for (int i = 0; i < n; i++)
    {
        if (A[i] == B[0])
            dp[i][0] = 1;
    }
    for (int j = 1; j < m; j++)
    {
        if (B[j] == A[0])
            dp[0][j] = 1;
    }
    for (int i = 1; i < n; i++)
    {
        for (int j = 1; j < m; j++)
        {
            if (A[i] == B[j])
                max_result = max(dp[i][j] = dp[i - 1][j - 1] + 1, max_result);
        }
    }
    return max_result;
}
int main()
{
    string A = "1AB2345CD";
    string B = "12345EF";
    int n = 9;
    int m = 7;
    int result = findLongest(A, n, B, m);
    cout << result << endl;
    return 0;
}

• 最小編輯距離，dp方法求解

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int findMinCost(string A, int n, string B, int m, int c0, int c1, int c2)
{
    int min_result = INT_MAX;
    vector< vector<int> > dp(n + 1, vector<int> (m + 1, 0));
    // 初始化dp
    for (int i = 1; i < m + 1; i++)
    {
        dp[0][i] += dp[0][i - 1] + c0;
    }
    for (int j = 1; j < n + 1; j++)
    {
        dp[j][0] += dp[j - 1][0] + c1;
    }
    // 中間動態規劃求解
    for (int i = 1; i < n + 1; i++)
    {
        for (int j = 1; j < m + 1; j++)
        {
            if (A[i - 1] == B[j - 1])
                dp[i][j] = dp[i - 1][j - 1];
            else
                dp[i][j] = dp[i - 1][j - 1] + c2;
            dp[i][j] = min(dp[i][j], min(dp[i][j - 1] + c0, dp[i - 1][j] + c1));
        }
    }
    return dp[n][m];
}
int main()
{
    string A = "abc";
    int n = 3;
    string B = "adc";
    int m = 3;
    int c0 = 5;
    int c1 = 3;
    int c2 = 100;
    int result = findMinCost(A, n, B, m, c0, c1, c2);
    cout << result << endl;
    return 0;
}

• 最長迴文子串

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int getLongestPalindrome(string A, int n)
{
    int maxlen = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = 1; j < n - i + 1; j++)
        {
            string temp = A.substr(i, j);
            reverse(temp.begin(), temp.end());
            if (temp == A.substr(i, j))
            {
                if (maxlen < j)
                    maxlen = j;
            }
        }
    }
    return maxlen;
}
int main()
{
    string A = "abc1234321ab";
    int n = 12;
    int result = getLongestPalindrome(A, n);
    cout << result << endl;
    return 0;
}

• 求解模式匹配演算法

#include <iostream>
#include <string>
#include <vector>
using namespace std;
int findAppearance1(string A, int lena, string B, int lenb)
{
    int i = 0;
    int j = 0;
    while (j < B.length() && i < A.length())
    {
        if (A[i] == B[j])
        {
            i++;
            j++;
        }
        else
        {
            i = i - j + 1;
            j = 0;
        }
    }
    if (j == B.length())
        return i - j;
    return -1;
}
void GextNextVal(string B, vector<int>& next)
{
    int nlen = B.length();
    next[0] = -1;
    int k = -1;
    int j = 0;
    while (j < nlen - 1)
    {
        if (k == -1 || B[j] == B[k])
        {
            j++;
            k++;
            if (B[j] != B[k])
                next[j] = k;
            else
                next[j] = next[k];
        }
        else
            k = next[k];
    }
}
int findAppearance(string A, int lena, string B, int lenb)
{
    int i = 0;
    int j = 0;
    vector<int> next(lenb, 0);
    GextNextVal(B, next);
    while (i < lena && j < lenb)
    {
        if (j == -1 || A[i] == B[j])
        {
            i++;
            j++;
        }
        else
        {
            j = next[j];
        }
        if (j == lenb)
            return i - j;
    }
    return -1;
}
int main()
{
    string A = "acbc";
    int n = 4;
    string B = "bc";
    int m = 2;
    int result = findAppearance(A, n, B, m);
    cout << result << endl;
    return 0;
}

• 字串轉整數

int myAtoi(char* str) 
{
    if(str == NULL)
        return 0;  
    bool sign = true;  
    while (*str == ' ')   
        str++;    
    if(*str == '+' || *str == '-')  
    {  
        if(*str == '+') 
            sign = true;  
        else 
            sign = false;  
        str++;  
    }   
    long long ans = 0;  
    while(*str >= '0' && *str <= '9')  
    {  
        ans = ans*10 + (*str - '0');  
        if(ans > INT_MAX) 
            return sign == true ? INT_MAX : INT_MIN;  
        str++;  
    }  
    if(sign == false) 
        return (int)(-ans);  
    else 
        return (int)ans;  
}