LWC 55：712. Minimum ASCII Delete Sum for Two Strings
LWC 55：712. Minimum ASCII Delete Sum for Two Strings
LWC 55：712. Minimum ASCII Delete Sum for Two Strings
传送门：712. Minimum ASCII Delete Sum for Two Strings
Problem:
Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.
Example 1:
Input: s1 = “sea”, s2 = “eat” Output: 231 Explanation: Deleting “s” from “sea” adds the ASCII value of “s” (115) to the sum. Deleting “t” from “eat” adds 116 to the sum. At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.
Example 2:
Input: s1 = “delete”, s2 = “leet” Output: 403 Explanation: Deleting “dee” from “delete” to turn the string into “let”, adds 100[d]+101[e]+101[e] to the sum. Deleting “e” from “leet” adds 101[e] to the sum. At the end, both strings are equal to “let”, and the answer is 100+101+101+101 = 403. If instead we turned both strings into “lee” or “eet”, we would get answers of 433 or 417, which are higher.
Note:
0 < s1.length, s2.length <= 1000.
All elements of each string will have an ASCII value in [97, 122].
思路： 动态规划，和最小编辑距离相似，先用递归的方式解一遍，定义f(i, j)表示：字符串（0，i）和字符串（0，j）该问题的解。
目标在于缩小子问题的规模，很简单，如果末尾的i和j对应的元素不相等，则一定删除其中的某个，所以有：
f(i, j) = min{f(i - 1, j) + 删除i的代价, f(i, j - 1) + 删除j的代价}
如果相等，则可删除可不删除。
f(i, j) = f(i - 1, j - 1)，但既然能不删，则删了岂不是答案更大，所以只要上述递推式即可。
递归+记忆化版本如下：
public int minimumDeleteSum(String s1, String s2) {
dp = new int[s1.length() + 1][s2.length() + 1];
return go(s1.toCharArray(), s2.toCharArray(), s1.length() - 1, s2.length() - 1);
}
int[][] dp;
int go(char[] cs1, char[] cs2, int p1, int p2) {
if (dp[p1 + 1][p2 + 1] > 0) return dp[p1 + 1][p2 + 1];
if (p1 == -1 && p2 == -1) return 0;
if (p1 == -1 && p2 != -1) {
int ans = go(cs1, cs2, p1, p2 - 1);
ans += cs2[p2];
dp[0][p2 + 1] = ans;
return ans;
}
if (p1 != -1 && p2 == -1) {
int ans = go(cs1, cs2, p1 - 1, p2);
ans += cs1[p1];
dp[p1 + 1][0] = ans;
return ans;
}
if (cs1[p1] == cs2[p2]) {
int ans = go(cs1, cs2, p1 - 1, p2 - 1);
dp[p1 + 1][p2 + 1] = ans;
return ans;
}
else {
int a1 = go(cs1, cs2, p1 - 1, p2);
int a2 = go(cs1, cs2, p1, p2 - 1);
return dp[p1 + 1][p2 + 1] = Math.min(a1 + cs1[p1], a2 + cs2[p2]);
}
}有了递归版本，咱们来个递推的（动态规划）：
public int minimumDeleteSum(String s1, String s2) {
char[] cs1 = s1.toCharArray();
char[] cs2 = s2.toCharArray();
int n1 = cs1.length;
int n2 = cs2.length;
int[][] dp = new int[n1 + 1][n2 + 1];
for (int i = 0, t = 0; i < n2; ++i) {
t += cs2[i];
dp[0][i + 1] = t;
}
for (int i = 0, t = 0; i < n1; ++i) {
t += cs1[i];
dp[i + 1][0] = t;
}
for (int i = 1; i <= n1; ++i) {
for (int j = 1; j <= n2; ++j) {
if (cs1[i - 1] == cs2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
}
else {
dp[i][j] = Math.min(dp[i][j - 1] + cs2[j - 1], dp[i - 1][j] + cs1[i - 1]);
}
}
}
return dp[n1][n2];
}