[pull] master from youngyangyang04:master #104

Original file line number	Diff line number	Diff line change
Expand Up		@@ -362,6 +362,33 @@ function minDistance(word1: string, word2: string): number {
		};
		```

	C:

	```c
	int min(int num1, int num2, int num3) {
	return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1);
	}

	int minDistance(char * word1, char * word2){
	int dp[strlen(word1)+1][strlen(word2)+1];
	dp[0][0] = 0;
	for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
	for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;

	for (int i = 1; i <= strlen(word1); i++) {
	for (int j = 1; j <=strlen(word2); j++) {
	if (word1[i-1] == word2[j-1]) {
	dp[i][j] = dp[i-1][j-1];
	}
	else {
	dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1;
	}
	}
	}
	return dp[strlen(word1)][strlen(word2)];
	}
	```



		-----------------------
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11 changes: 7 additions & 4 deletions problems/0150.逆波兰表达式求值.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -89,26 +89,29 @@ C++代码如下:
		class Solution {
		public:
		int evalRPN(vector<string>& tokens) {
	stack<int> st;
	// 力扣修改了后台测试数据,需要用longlong
	stack<long long> st;
		for (int i = 0; i < tokens.size(); i++) {
		if (tokens[i] == "+" \|\| tokens[i] == "-" \|\| tokens[i] == "*" \|\| tokens[i] == "/") {
	int num1 = st.top();
	long long num1 = st.top();
		st.pop();
	int num2 = st.top();
	long long num2 = st.top();
		st.pop();
		if (tokens[i] == "+") st.push(num2 + num1);
		if (tokens[i] == "-") st.push(num2 - num1);
		if (tokens[i] == "") st.push((long)num2 (long)num1); //力扣改了后台测试数据
		if (tokens[i] == "/") st.push(num2 / num1);
		} else {
	st.push(stoi(tokens[i]));
	st.push(stoll(tokens[i]));
		}
		}

		int result = st.top();
		st.pop(); // 把栈里最后一个元素弹出(其实不弹出也没事)
		return result;
		}
		};

		```

		## 题外话
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4 changes: 3 additions & 1 deletion problems/0583.两个字符串的删除操作.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -47,6 +47,8 @@ dp[i][j]:以i-1为结尾的字符串word1,和以j-1位结尾的字符串word

		那最后当然是取最小值,所以当word1[i - 1] 与 word2[j - 1]不相同的时候,递推公式:dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});

	因为dp[i - 1][j - 1] + 1等于 dp[i - 1][j] 或 dp[i][j - 1],所以递推公式可简化为:dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);


		3. dp数组如何初始化

Expand Down Expand Up		@@ -90,7 +92,7 @@ public:
		if (word1[i - 1] == word2[j - 1]) {
		dp[i][j] = dp[i - 1][j - 1];
		} else {
	dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
	dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
		}
		}
		}
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