题目描述：给定两个字符串s1, s2，找到使两个字符串相等所需删除字符的ASCII值的最小和。
代码实现
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| public class Node {
private int seqLen;
private int sumAsc;
private String longestSeq;
public int getSeqLen() {
return seqLen;
}
public void setSeqLen(int seqLen) {
this.seqLen = seqLen;
}
public String getLongestSeq() {
return longestSeq;
}
public void setLongestSeq(String longestSeq) {
this.longestSeq = longestSeq;
}
public int getSumAsc() {
return sumAsc;
}
public void setSumAsc(int sumAsc) {
this.sumAsc = sumAsc;
}
public Node(int seqLen, int sumAsc, String longestSeq) {
this.seqLen = seqLen;
this.sumAsc = sumAsc;
this.longestSeq = longestSeq;
}
}
|
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| class Solution {
public int minimumDeleteSum(String s1, String s2) {
Node maxNode = longestCommonSubsequence(s1.toCharArray(), s2.toCharArray());
int sumS1 = 0;
for (int i = 0; i < s1.length(); i++) {
sumS1 += s1.charAt(i);
}
int sumS2 = 0;
for (int i = 0; i < s2.length(); i++) {
sumS2 += s2.charAt(i);
}
int sumM = 0;
for (int i = 0; i < maxNode.getLongestSeq().length(); i++) {
sumM += maxNode.getLongestSeq().charAt(i);
}
return sumS1 + sumS2 - sumM * 2;
}
private Node longestCommonSubsequence(char[] nums1, char nums2[]) {
Node[][] dp = new Node[nums1.length + 1][nums2.length + 1];
for (int i = 0; i <= nums1.length; i++) {
for (int j = 0; j <= nums2.length; j++) {
dp[i][j] = new Node(0, 0, "");
}
}
for (int i = 1; i <= nums1.length; i++) {
for (int j = 1; j <= nums2.length; j++) {
if (nums1[i - 1] == nums2[j - 1]) {
dp[i][j].setSeqLen(dp[i - 1][j - 1].getSeqLen() + 1);
dp[i][j].setLongestSeq(dp[i - 1][j - 1].getLongestSeq() + nums1[i - 1]);
dp[i][j].setSumAsc(dp[i - 1][j - 1].getSumAsc() + nums1[i - 1]);
} else {
Node rNode = dp[i - 1][j];
Node cNode = dp[i][j - 1];
Node maxNode = null;
if (rNode.getSeqLen() == cNode.getSeqLen()) {
maxNode = rNode.getSumAsc() > cNode.getSumAsc() ? rNode : cNode;
} else if (rNode.getSeqLen() < cNode.getSeqLen()) {
maxNode = cNode;
} else {
maxNode = rNode;
}
dp[i][j] = maxNode;
}
}
}
return dp[nums1.length][nums2.length];
}
}
|