1. 编辑距离

给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数。

你可以对一个单词进行如下三种操作：

插入一个字符
删除一个字符
替换一个字符

示例 1：

输入：word1 = "horse", word2 = "ros"
输出：3
解释：horse -> rorse (将 'h' 替换为 'r')rorse -> rose (删除 'r')rose -> ros (删除 'e')

示例 2：

输入：word1 = "intention", word2 = "execution"
输出：5
解释：intention -> inention (删除 't')inention -> enention (将 'i' 替换为 'e')enention -> exention (将 'n' 替换为 'x')exention -> exection (将 'n' 替换为 'c')exection -> execution (插入 'u')

提示：

0 <= word1.length, word2.length <= 500
word1 和 word2 由小写英文字母组成

以下程序实现了这一功能，请你填补空白处的内容：

```c++
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int minDistance(string word1, string word2)
{
int l1 = word1.length();
int l2 = word2.length();
vector<int> dp(l2 + 1);
for (int i = 0; i <= l2; i++)
{
dp[i] = i;
}
int up = 0;
for (int i = 1; i <= l1; i++)
{
int left_up = dp[0];
dp[0] = i;
for (int j = 1; j <= l2; j++)
{
up = dp[j];
_________________________;
else
{
dp[j] = 1 + min(left_up, min(up, dp[j - 1]));
}
left_up = up;
}
}
return dp[l2];
}
};
```

出处：

https://edu.csdn.net/practice/27452676

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    int minDistance(string word1, string word2)
    {
        int l1 = word1.length();
        int l2 = word2.length();
        vector<int> dp(l2 + 1);
        for (int i = 0; i <= l2; i++)
        {
            dp[i] = i;
        }
        int up = 0;
        for (int i = 1; i <= l1; i++)
        {
            int left_up = dp[0];
            dp[0] = i;
            for (int j = 1; j <= l2; j++)
            {
                up = dp[j];
                if (word1[i - 1] == word2[j - 1])
                {
                    dp[j] = left_up;
                }
                else
                {
                    dp[j] = 1 + min(left_up, min(up, dp[j - 1]));
                }
                left_up = up;
            }
        }
        return dp[l2];
    }
};

int main()
{
	Solution s;
	string word1 = "horse";
	string word2 = "ros";
    cout << s.minDistance(word1, word2) << endl;
    word1 = "intention";
	word2 = "execution";
	cout << s.minDistance(word1, word2) << endl;
	
	return 0;
}

输出：

3
5

2. 多数元素

给定一个大小为 n 的数组，找到其中的多数元素。多数元素是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。

你可以假设数组是非空的，并且给定的数组总是存在多数元素。

示例 1：

输入：[3,2,3]
输出：3

示例 2：

输入：[2,2,1,1,1,2,2]
输出：2

进阶：

尝试设计时间复杂度为 O(n)、空间复杂度为 O(1) 的算法解决此问题。

出处：

https://edu.csdn.net/practice/27452677

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    int majorityElement(vector<int> &nums)
    {
        unordered_map<int, int> counts;
        int majority = 0, cnt = 0;
        for (int num : nums)
        {
            ++counts[num];
            if (counts[num] > cnt)
            {
                majority = num;
                cnt = counts[num];
            }
        }
        return majority;
    }
};

int main()
{
	Solution s;
	vector<int> nums = {3,2,3};
    cout << s.majorityElement(nums) << endl;
    nums = {2,2,1,1,1,2,2};
	cout << s.majorityElement(nums) << endl;
	
	return 0;
}

输出：

3
2

3. 求分数数列的前N项和

有一分数序列:2/1,-3/2,5/3,-8/5,13/8,-21/13,…, 由用户输入项目数N，求这个数列的前N 项之和

以下程序实现了这一功能，请你填补空白处内容：

```c++
#include <stdlib.h>
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int i;
double a1 = 2, b1 = 1;
double a2 = 3, b2 = 2;
double sum = a1 / b1 - a2 / b2;
if (n == 1)
printf("%f\n", a1 / b1);
else if (n == 2)
printf("%f\n", sum);
else
{
for (i = 0; i < n - 2; i++)
{
double exp = a2 / b2;
_____________________;
sum += exp;
double a = a1 + a2;
double b = b1 + b2;
a1 = a2;
b1 = b2;
a2 = a;
b2 = b;
}
printf("%f\n", sum);
}
return 0;
}
```

出处：

https://edu.csdn.net/practice/27452678

代码：

#include <stdlib.h>
#include <stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    int i;
    double a1 = 2, b1 = 1;
    double a2 = 3, b2 = 2;
    double sum = a1 / b1 - a2 / b2;
    if (n == 1)
        printf("%f\n", a1 / b1);
    else if (n == 2)
        printf("%f\n", sum);
    else
    {
        for (i = 0; i < n - 2; i++)
        {
            double exp = a2 / b2;
            if (i % 2 == 0)
                exp *= -1;
            sum += exp;
            double a = a1 + a2;
            double b = b1 + b2;
            a1 = a2;
            b1 = b2;
            a2 = a;
            b2 = b;
        }
        printf("%f\n", sum);
    }
    return 0;
}

输出：

6
0.691667