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LeetCode-Interleaving String[dp]
Interleaving String
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc"
,
s2 = "dbbca"
,
When s3 = "aadbbcbcac"
, return true.
When s3 = "aadbbbaccc"
, return false.
标签: Dynamic Programming String
分析:动态规划;设boolean数组dp[i][j]表示字符串s1[0...i-1]和字符串s2[0...j-1]是否可以匹配到字符串s3[0...i+j-1],
因此有字符s3[i+j-1]可以由字符s1[i-1]或字符s2[j-1]来匹配,所以:
如果s1[i-1]==s3[i+j-1],则dp[i][j]=dp[i-1][j];
如果s2[j-1]==s3[i+j-1],则dp[i][j]=dp[i][j-1];
所以状态转移方程为:dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1])
参考代码:
public class Solution { public boolean isInterleave(String s1, String s2, String s3) { int len1=s1.length(); int len2=s2.length(); int len3=s3.length(); if(len1+len2!=len3) return false; boolean dp[][]=new boolean[len1+1][len2+1]; dp[0][0]=true; for(int j=1;j<=len2;j++){ dp[0][j]=dp[0][j-1]&&s2.charAt(j-1)==s3.charAt(j-1); } for(int i=1;i<=len1;i++){ dp[i][0]=dp[i-1][0]&&s1.charAt(i-1)==s3.charAt(i-1); } for(int i=1;i<=len1;i++){ for(int j=1;j<=len2;j++){ dp[i][j]=(dp[i-1][j]&&s1.charAt(i-1)==s3.charAt(i+j-1))||(dp[i][j-1]&&s2.charAt(j-1)==s3.charAt(i+j-1)); } } return dp[len1][len2]; } }
LeetCode-Interleaving String[dp]
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