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 貪心  二分查找。

1.最簡單的作法，掃描一遍就好

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. 暴力遞歸搜索

bool f(int a, int b)
{
	if (sa =="" )return true;

	if (a > sa.size())return true;//若是大於 那麼表示前面全部的都匹配了所以true
 
	if (b > sb.size())return false;//b的index大於大小，意味着搜索完了還沒return true所以false

	if (sa[a] == sb[b])//若是相同 那麼直接遞歸下一個index
	{
		return f(a + 1, b + 1);
	}
	else
	{
		return    f(a, b + 1);//不一樣則移動sb的index
	}
}

 

3.吧暴力遞歸轉換爲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的狀況
	}
	for (int a =  sa.size() - 1;a>=0; a--)//遞推求解，遞推公式和遞歸公式同樣，可是循環條件是逆向，由於遞歸到底之後是倒着返回來的，因此遞推要倒着
	{
		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;