代码随想录 Day54 - 动态规划(十五)
作者:jjn0703
- 2023-08-26 江苏
本文字数:1000 字
阅读完需:约 3 分钟
作业题
392.判断子序列
链接:https://leetcode.cn/problems/is-subsequence/description/
package jjn.carl.dp;
import java.util.Scanner;
/** * @author Jjn * @since 2023/8/26 11:06 */public class LeetCode392 { public boolean isSubsequence(String s, String t) { int firstIndex = 0; int secondIndex = 0; while (secondIndex < t.length() && firstIndex < s.length()) { if (t.charAt(secondIndex) == s.charAt(firstIndex)) { firstIndex++; } secondIndex++; } return firstIndex >= s.length(); } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String s = scanner.next(); String t = scanner.next(); boolean isSubsequence = new LeetCode392().isSubsequence(s, t); System.out.println(isSubsequence); }}
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115.不同的子序列
链接:https://leetcode.cn/problems/distinct-subsequences/description/
package jjn.carl.dp;
import jjn.round1.LeetCode171_ExcelSheetColumnNumber;
import java.util.Scanner;
/** * @author Jjn * @since 2023/8/26 11:29 */public class LeetCode115 { public int numDistinct(String s, String t) { int[][] dp = new int[s.length() + 1][t.length() + 1]; for (int i = 0; i <= s.length(); i++) { dp[i][0] = 1; } for (int i = 1; i <= s.length(); i++) { for (int j = 1; j <= t.length(); j++) { if (s.charAt(i - 1) == t.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; }else{ dp[i][j] = dp[i - 1][j]; } } } return dp[s.length()][t.length()]; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String s = scanner.next(); String t = scanner.next(); int distinct = new LeetCode115().numDistinct(s, t); System.out.println(distinct); }}
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版权声明: 本文为 InfoQ 作者【jjn0703】的原创文章。
原文链接:【http://xie.infoq.cn/article/af10855963e99a4cf8691cdfa】。
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