|
|
@@ -0,0 +1,64 @@
|
|
|
+#include <string>
|
|
|
+
|
|
|
+using std::string;
|
|
|
+
|
|
|
+/* BEGIN */
|
|
|
+
|
|
|
+#include <cstring>
|
|
|
+#include <algorithm>
|
|
|
+
|
|
|
+using std::min;
|
|
|
+
|
|
|
+#define L 1000
|
|
|
+#define INF 1e8
|
|
|
+
|
|
|
+int dp[L + 1][L + 1];
|
|
|
+
|
|
|
+class Solution {
|
|
|
+public:
|
|
|
+ int minimumDeleteSum(string s1, string s2) {
|
|
|
+ memset(dp, 0, sizeof(dp));
|
|
|
+ for (int i = 1; i <= s1.size(); i++) { // The boundary is very important!
|
|
|
+ dp[i][0] = dp[i - 1][0] + s1.at(i - 1);
|
|
|
+ }
|
|
|
+ for (int j = 1; j <= s2.size(); j++) {
|
|
|
+ dp[0][j] = dp[0][j - 1] + s2.at(j - 1);
|
|
|
+ }
|
|
|
+ for (int i = 1; i <= s1.size(); i++) {
|
|
|
+ for (int j = 1; j <= s2.size(); j++) {
|
|
|
+ int c1 = s1.at(i - 1), c2 = s2.at(j - 1);
|
|
|
+ int cost = min(dp[i - 1][j] + c1, dp[i][j - 1] + c2);
|
|
|
+ if (c1 == c2) {
|
|
|
+ dp[i][j] = min(dp[i - 1][j - 1], cost);
|
|
|
+ } else {
|
|
|
+ dp[i][j] = min(dp[i - 1][j - 1] + c1 + c2, cost);
|
|
|
+ }
|
|
|
+ }
|
|
|
+ }
|
|
|
+ return dp[s1.size()][s2.size()];
|
|
|
+ }
|
|
|
+};
|
|
|
+
|
|
|
+/* END */
|
|
|
+
|
|
|
+#include <iostream>
|
|
|
+
|
|
|
+using std::cout; using std::endl;
|
|
|
+
|
|
|
+int main() {
|
|
|
+ string s1 = "delete";
|
|
|
+ string s2 = "leet";
|
|
|
+ Solution *sol = new Solution();
|
|
|
+ int ans = sol->minimumDeleteSum(s1, s2);
|
|
|
+ cout << s1 << " vs " << s2 << ": " << ans << endl;
|
|
|
+
|
|
|
+ s1 = "sea";
|
|
|
+ s2 = "eat";
|
|
|
+ ans = sol->minimumDeleteSum(s1, s2);
|
|
|
+ cout << s1 << " vs " << s2 << ": " << ans << endl;
|
|
|
+
|
|
|
+ s1 = "a";
|
|
|
+ s2 = "at";
|
|
|
+ ans = sol->minimumDeleteSum(s1, s2);
|
|
|
+ cout << s1 << " vs " << s2 << ": " << ans << endl;
|
|
|
+}
|
