leetcode-392. Is Subsequence-DP-NORMAL
Given a string s and a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).
Example 1:
s = "abc", t = "ahbgdc"
Return true.
Example 2:
s = "axc", t = "ahbgdc"
Return false.
Follow up:
If there are lots of incoming S, say S1, S2, … , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?
可用DP 贪心 二分查找。
1.最简单的做法，扫描一遍就好
if (s == "")return true;
int index = 0;
for (int i = 0; i < t.size(); i++)
{
if (s[index] == t[i])
{
index++;
if (index >= s.size())return true;
}
}
return false;
2. 暴力递归搜索
bool f(int a, int b)
{
if (sa =="" )return true;
if (a > sa.size())return true;//如果大于 那么表示前面所有的都匹配了因此true
if (b > sb.size())return false;//b的index大于大小，意味着搜索完了还没return true因此false
if (sa[a] == sb[b])//如果相同 那么直接递归下一个index
{
return f(a + 1, b + 1);
}
else
{
return f(a, b + 1);//不同则移动sb的index
}
}
3.吧暴力递归转换为DP
bool arr[10][10];//a b
memset(arr, 0, 10 * 10);
for (int b = 0; b < 10;b++)arr[0][b] = true;//初始化 字符串空
for (int a = sa.size(); a < 10; a++)
{
for (int b =0; b < 10; b++)
arr[a][b] = true; // 初始化a>sa.size的情况
}
for (int a = sa.size() - 1;a>=0; a--)//递推求解，递推公式和递归公式一样，但是循环条件是逆向，因为递归到底以后是倒着返回来的，所以递推要倒着
{
for (int b = sb.size() - 1;b>=0; b--)
{
if (sa[a] == sb[b])
{
arr[a][b] = arr[a + 1][b + 1];
}
else
{
arr[a][b] = arr[a][b + 1];
}
}
}
cout << arr[0][0] << endl;
cout << f(0, 0) << endl;