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Merge pull request #1670 from chenzhg-maker/master
Update 0583.两个字符串的删除操作 C++
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@ -362,6 +362,33 @@ function minDistance(word1: string, word2: string): number {
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};
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```
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C:
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```c
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int min(int num1, int num2, int num3) {
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return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1);
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}
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int minDistance(char * word1, char * word2){
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int dp[strlen(word1)+1][strlen(word2)+1];
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dp[0][0] = 0;
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for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
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for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;
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for (int i = 1; i <= strlen(word1); i++) {
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for (int j = 1; j <=strlen(word2); j++) {
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if (word1[i-1] == word2[j-1]) {
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dp[i][j] = dp[i-1][j-1];
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}
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else {
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dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1;
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}
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}
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}
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return dp[strlen(word1)][strlen(word2)];
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}
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```
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-----------------------
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@ -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
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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});
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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);
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3. dp数组如何初始化
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@ -90,7 +92,7 @@ public:
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if (word1[i - 1] == word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
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dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
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}
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}
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}
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