最长公共字符串

最长公共字符串

// dp + 空间优化（重要）
// DC3后缀串解法了解

// dp  
public static int lcsM1(String s1, String s2) {  
	if (s1 == null || s1.length() == 0 || s2 == null || s2.length() == 0) {  
		return 0;  
	}  
	char[] str1 = s1.toCharArray();  
	char[] str2 = s2.toCharArray();  
	int N = str1.length, M = str2.length;  
	// dp[i][j]: str1[0~i]str2[0~j]严格已i,j结尾的最大公共子串  
	// 状态转移：  
	// i = 0; dp[0][j] = str1[0] == str2[j] ? 1 : 0;  
	// j = 0; dp[i][0] = str2[0] == str1[i] ? 1 : 0;  
	// i > 0 && j > 0:  
	// 1) str1[i] != str2[j] -> dp[i][j] = 0;  
	// 2) str1[i] == str2[j] -> dp[i][j] = dp[i - 1][j - 1] + 1;  
	int[][] dp = new int[N][M];  
	dp[0][0] = str1[0] == str2[0] ? 1 : 0;  
	for (int i = 0; i < N; i++) {  
		dp[i][0] = str1[i] == str2[0] ? 1 : 0;  
	}  
	for (int j = 0; j < M; j++) {  
		dp[0][j] = str2[j] == str1[0] ? 1 : 0;  
	}  
	for (int i = 1; i < N; i++) {  
		for (int j = 1; j < M; j++) {  
			if (str1[i] == str2[j]) {  
				dp[i][j] = dp[i - 1][j - 1] + 1;  
			}  
		}  
	}  
	int ans = Integer.MIN_VALUE;  
	for (int i = 0; i < N; i++) {  
		for (int j = 0; j < M; j++) {  
			ans = Math.max(ans, dp[i][j]);  
		}  
	}  
	return ans;  
}  
  
// dp + 空间压缩  
// 从lcs1的dp[i][j]状态转移仅依赖dp[i - 1][j - 1];  
// 逻辑上可以写在填表, 优化物理dp表空间  
// 1334  
// 012346  
public static int lcsM2(String s1, String s2) {  
	if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) {  
		return 0;  
	}  
	char[] str1 = s1.toCharArray(), str2 = s2.toCharArray();  
	int N = str1.length, M = str2.length;  
	int row = N - 1, col = M - 1;  
	int ans = 0;  
	while (row >= 0) {  
		int len = 0;  
		for (int j = 0, i = row; i < N && j < M; i++, j++) {  
			len = str1[i] == str2[j] ? len + 1 : 0;  
			ans = Math.max(ans, len);  
		}  
		row--;  
	}  
	while (col >= 0) {  
		int len = 0;  
		for (int i = 0, j = col; i < N && j < M; i++, j++) {  
			len = str1[i] == str2[j] ? len + 1 : 0;  
			ans = Math.max(ans, len);  
		}  
		col--;  
	}  
	return ans + 1;  
}