两个求解代码类似的题目,对比记忆!!!
1143.最长公共子序列
法1:DP,考虑空串
java
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length() + 1, n = text2.length() + 1; // 加1是在构建二维矩阵时增加空串情况, 简化计算
int[][] dp = new int[m][n];
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
return dp[m - 1][n - 1];
}
}
法2:DP,不考虑空串
代码会啰嗦很多,所以还是法1好哇!!!
java
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length(), n = text2.length();
int[][] dp = new int[m][n];
boolean existed = false;
for (int i = 0; i < n; ++i) {
dp[0][i] = (existed || text1.charAt(0) == text2.charAt(i)) ? 1 : 0;
if (dp[0][i] == 1) {
existed = true;
}
}
existed = false;
for (int i = 0; i < m; ++i) {
dp[i][0] = (existed || text1.charAt(i) == text2.charAt(0)) ? 1 : 0;
if (dp[i][0] == 1) {
existed = true;
}
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (text1.charAt(i) == text2.charAt(j)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
return dp[m - 1][n - 1];
}
}
516.最长回文子序列
java
class Solution {
public int longestPalindromeSubseq(String s) {
int n = s.length();
int[][] dp = new int[n][n];
for (int i = 0; i < n; ++i) {
dp[i][i] = 1;
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = (j - i == 1 ? 0 : dp[i + 1][j - 1]) + 2;
} else {
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
return dp[0][n - 1];
}
}