[Leetcode] 97. Interleaving String 解题报告
2017-04-20 11:15
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题目:
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 =
s2 =
When s3 =
When s3 =
思路:
又是一道典型的动态规划题目:我们定义dp[i][j]表示s1的前i个字符和s2的前j个字符是否可以以interleaving的方式形成s3的前(i+j)个字符。那么递推式可以分为如下两种情况:
1)如果s1[i-1] == s3[i+j-1],此时检查dp[i-1][j]是否为true即可;
2)如果s2[j-1] == s3[i+j-1],此时检查dp[i][j-1]是否为true即可。
以上两个条件只要满足其一,就说明s1的前i个字符和s2的前j个字符是否可以以interleaving的方式形成s3的前(i+j)个字符。
代码:
class Solution {
public:
bool isInterleave(string s1, string s2, string s3)
{
if(s1.length() + s2.length() != s3.length()) {
return false;
}
vector<vector<bool>> dp(s1.length() + 1, vector<bool>(s2.length() + 1, true));
dp[0][0] = true;
for(int i = 1; i <= s1.length(); ++i) {
dp[i][0] = (s1[i - 1] == s3[i - 1] && dp[i-1][0]);
}
for(int j = 1; j <= s2.length(); ++j) {
dp[0][j] = (s2[j-1] == s3[j-1] && dp[0][j-1]);
}
for(int i = 1; i <= s1.length(); ++i) {
for(int j = 1; j <= s2.length(); ++j) {
dp[i][j] = (s1[i-1] == s3[i+j-1] && dp[i-1][j]) || (s2[j-1] == s3[i+j-1] && dp[i][j-1]);
}
}
return dp[s1.length()][s2.length()];
}
};
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.
思路:
又是一道典型的动态规划题目:我们定义dp[i][j]表示s1的前i个字符和s2的前j个字符是否可以以interleaving的方式形成s3的前(i+j)个字符。那么递推式可以分为如下两种情况:
1)如果s1[i-1] == s3[i+j-1],此时检查dp[i-1][j]是否为true即可;
2)如果s2[j-1] == s3[i+j-1],此时检查dp[i][j-1]是否为true即可。
以上两个条件只要满足其一,就说明s1的前i个字符和s2的前j个字符是否可以以interleaving的方式形成s3的前(i+j)个字符。
代码:
class Solution {
public:
bool isInterleave(string s1, string s2, string s3)
{
if(s1.length() + s2.length() != s3.length()) {
return false;
}
vector<vector<bool>> dp(s1.length() + 1, vector<bool>(s2.length() + 1, true));
dp[0][0] = true;
for(int i = 1; i <= s1.length(); ++i) {
dp[i][0] = (s1[i - 1] == s3[i - 1] && dp[i-1][0]);
}
for(int j = 1; j <= s2.length(); ++j) {
dp[0][j] = (s2[j-1] == s3[j-1] && dp[0][j-1]);
}
for(int i = 1; i <= s1.length(); ++i) {
for(int j = 1; j <= s2.length(); ++j) {
dp[i][j] = (s1[i-1] == s3[i+j-1] && dp[i-1][j]) || (s2[j-1] == s3[i+j-1] && dp[i][j-1]);
}
}
return dp[s1.length()][s2.length()];
}
};
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