java暴力匹配及KMP算法解决字符串匹配问题示例详解

String str1 = "BBC ABCDAB ABCDABCDABDE";
String str2 = "ABCDABD";

public class ViolenceMatch {
	public static void main(String[] args) {
		String str1 = "硅硅谷 尚硅谷你尚硅 尚硅谷你尚硅谷你尚硅你好";
		String str2 = "尚硅谷你尚硅你";
		int index = violenceMatch(str1, str2);
		System.out.println("index=" + index);
	}
	
	public static int violenceMatch(String str1, String str2) {
		char[] s1 = str1.toCharArray();
		char[] s2 = str2.toCharArray();
 
		int s1Len = s1.length;
		int s2Len = s2.length;
 
		int i = 0;// i索引指向s1
		int j = 0;// j索引指向s2
		while (i < s1Len && j < s2Len) {// 保证匹配时，不越界
			if (s1[i] == s2[j]) {// 匹配ok
				i++;
				j++;
			} else {// 没有匹配成功
					// 如果不匹配(即str1[i] != str2[j],令i = i - (j - 1),j = 0)
				i = i - (j - 1);
				j = 0;
			}
		}
		// 判断是否匹配成功
		if (j == s2Len) {
			return i - j;
		} else {
			return -1;
		}
	}
}

public class KMPAlGorithm {
	public static void main(String[] args) {
		String str1 = "BBC ABCDAB ABCDABCDABDE";
		String str2 = "ABCDABD";
		int[] next = kmpNext("ABCDABD");
		System.out.println("next=" + Arrays.toString(next));
 		int index = kmpSearch(str1, str2, next);
		System.out.println("index=" + index);
	} 
	
	public static int kmpSearch(String str1, String str2, int[] next) {
		// 遍历
		for (int i = 0, j = 0; i < str1.length(); i++) {
 
			// 需要处理 str1.charAt(i) != str2.charAt(j),去调整j的大小
			// KMP算法核心点
			while (j > 0 && str1.charAt(i) != str2.charAt(j)) {
				j = next[j - 1];
			}
			if (str1.charAt(i) == str2.charAt(j)) {
				j++;
			}
			if (j == str2.length()) {// 找到了
				return i - j + 1;
			}
		}
		return -1;
	} 
	
	public static int[] kmpNext(String dest) {
		// 创建一个next数组保存部分匹配值
		int[] next = new int[dest.length()];
		next[0] = 0;// 如果字符串是长度为1部分匹配值就是0
		for (int i = 1, j = 0; i < dest.length(); i++) {
			// 当dest.charAt(i) != dest.charAt(j)，需要从next[j - 1]获取新的j
			// 直到发现有dest.charAt(i) == dest.charAt(j)成立才退出
			// 这是kmp算法核心点
			while (j > 0 && dest.charAt(i) != dest.charAt(j)) {
				j = next[j - 1];
			}
			if (dest.charAt(i) == dest.charAt(j)) {
				j++;
			}
			next[i] = j;
		}
		return next;
	}
}