NYOJ 题目5 字符串匹配

Binary String Matching

时间限制：65535 KB

难度：3

 

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

 

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

字符串查找有很多解法， 这里我分享下我的思路：

	public class StringFind {
		public static void main(String [] args) throws IOException {
		Scanner input = new Scanner(System.in);
			int n = input.nextInt();	
			for(int i=0; i<n; i++) {
				String sub = input.next();   // 要查找的子串
				String src = input.next();   // 目标字符串
				char[] subStr = sub.toCharArray();  
				char[] srcStr = src.toCharArray();
				ArrayList<Integer> list = new ArrayList<Integer>();
				ArrayList<Integer> temp = new ArrayList<Integer>();
				for(int j=0; j<srcStr.length-subStr.length+1; j++) { // 第一次循环次数
					list.add(j);
				}
				for(int j=0; j<subStr.length; j++) {  // 总共查找次数为字串的字符数
					//System.out.print("\n第"+j+"次：");
					temp.clear();
					for(int k=0; k<list.size(); k++) {
						int pos = list.get(k);
						if(srcStr[pos] == subStr[j]) {
							//System.out.print(pos+" ");
							int z = pos+1;
							temp.add(z);  //每次的下一个字符符合， 则将该下标存入tmp 中。
						}
					}
					list.clear();
					list.addAll(temp);
					//System.out.println("list.size(): " + list.size());
				}
				//System.out.print("\n 剩余: ");
				//for(int j=0; j<list.size(); j++) {
					//System.out.print(list.get(j)+" ");
				//}
				// 表中个数即父串中字串符合个数
				System.out.println(list.size());
			}
		}
	}