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C语言之运算符用法(补充前面运算符中的不足)

目录

LeeCode 583. 两个字符串的删除操作 

LeeCode 72. 编辑距离 


LeeCode 583. 两个字符串的删除操作 

583. 两个字符串的删除操作 - 力扣(LeetCode)

动规五部曲:

1.确定dp数组及下标含义: 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word2达到相等所需要删除元素的最少次数;

2.确定递推公式:if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1]; 

                            else  dp[i][j] = dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);

3.dp数组初始化:dp[i][0] = i; dp[0][j] = j;

4.确定遍历顺序:从左到右、从上到下遍历;

5.举例递推dp数组;

代码

class Solution {
public:
    int minDistance(string word1, string word2) {
    	vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
		for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
		for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
		for (int i = 1; i <= word1.size(); i++) {
			for (int j = 1; j <= word2.size(); 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] + 1);
			}
		} 
        return dp[word1.size()][word2.size()];
    }
};

时间复杂度:O(n*m)                                           空间复杂度:O(n*m)


LeeCode 72. 编辑距离 

72. 编辑距离 - 力扣(LeetCode)

动规五部曲:

1.确定dp数组及下标含义: dp[i][j]:以下标i-1为结尾的字符串word1和以下标j-1为结尾的字符串word2,最近编辑距离;

2.确定递推公式: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 - 1][j], dp[i][j - 1]}) + 1;

3.dp数组初始化:dp[i][0] = i; dp[0][j] = j;

4.确定遍历顺序:从左到右、从上到下遍历;

5.举例递推dp数组;

代码

class Solution {
public:
    int minDistance(string word1, string word2) {
    	vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
    	for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
    	for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
    	for (int i = 1; i <= word1.size(); i++) {
    		for (int j = 1; j <= word2.size(); 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 - 1][j], dp[i][j - 1]}) + 1;
			}
		}
        return dp[word1.size()][word2.size()];
    }
};

时间复杂度:O(n*m)                                           空间复杂度:O(n*m)

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