LeetCode[97]::Interleaving String C++
2015-08-22 23:40
459 查看
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 =
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int len1 = s1.length();
int len2 = s2.length();
int len3 = s3.length();
if( len1 == 0 && len2 == 0 && len3 == 0){
return true;
}
if(len3 != ( len1 + len2 )){
return false;
}
vector<bool> dp(len2+1);
for(size_t i = 0; i <= len1; ++i){
for(size_t j = 0; j <= len2; ++j){
if(i == 0){
if(j == 0){
dp[j] = true;
}else{
dp[j] = (dp[j-1] && (s2[j-1] == s3[i+j-1]));
}
}else{
if(j == 0){
dp[j] = (dp[j] && (s1[i-1] == s3[i+j-1]));
}else{
dp[j] = (dp[j] && s1[i-1] == s3[i+j-1]) || (dp[j-1] && s2[j-1] == s3[i+j-1]);
}
}
}
}
return dp[len2];
}
};
s1 and s2.
For example,
Given:
s1 =
"aabcc",
s2 =
"dbbca",
When s3 =
"aadbbcbcac", return true.
When s3 =
"aadbbbaccc", return false.
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int len1 = s1.length();
int len2 = s2.length();
int len3 = s3.length();
if( len1 == 0 && len2 == 0 && len3 == 0){
return true;
}
if(len3 != ( len1 + len2 )){
return false;
}
vector<bool> dp(len2+1);
for(size_t i = 0; i <= len1; ++i){
for(size_t j = 0; j <= len2; ++j){
if(i == 0){
if(j == 0){
dp[j] = true;
}else{
dp[j] = (dp[j-1] && (s2[j-1] == s3[i+j-1]));
}
}else{
if(j == 0){
dp[j] = (dp[j] && (s1[i-1] == s3[i+j-1]));
}else{
dp[j] = (dp[j] && s1[i-1] == s3[i+j-1]) || (dp[j-1] && s2[j-1] == s3[i+j-1]);
}
}
}
}
return dp[len2];
}
};
相关文章推荐
- c++:函数模板与函数包装器
- C语言字符串、指针和内存问题总结
- C++ Primer 5e chapter 13.2
- C语言打印九九乘法表
- [C语言]结构体
- [C语言]内存管理
- C语言undefined behaviour未定义行为
- C语言三种循环反汇编分析
- LeetCode[91]::Decode Ways C++
- C++ Primer 5e chapter 13.1
- C语言中的内存对齐
- C++中 cin 与 cout 的用法
- C/C++ 笔记(1)-- malloc 的工作原理
- c++中构造函数和析构函数执行过程
- c++中基类写成虚函数的作用
- C语言的取模结果
- C++中的智能指针笔记
- C++中const、volatile、mutable用法小结
- 内存管理——(exceptional C++ 条款9,条款10)
- C++程序的编译和运行