最长公共子串

发表于 | 分类于

给定两个字符串str1和str2，返回两个字符串的最长公共子串

解法一：动态规划

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
public int[][] getdp(char[] str1, char[] str2){
	int[][] dp = nex int[str1.length][str2.length];
	for (int i =0; i < str1.length; i++){
		if (str1[i] == str2[0]){
			dp[i][0] = 1;
		}
	}
	for (int j =0; j < str2.length; j++){
		if (str1[0] == str2[j]){
			dp[0][j] = 1;
		}
	}
	for (int i = 1; i < str1.length; i++){
		for (int j = 1; j < str2.length; j++){
			if(str1[i] == str2[j]){
				dp[i][j] = dp[i-1][j-1] +1;
			}
		}
	}
	return dp;
}

public String lcst1(String str1, String str2){
	if (str1 == null || str2 == null || str1 == "" || str2 == ""){
		return "";
	}
	char[] chs1 = str1.toCharArray();
	char[] chs2 = str2.tocharArray();
	
	int[][] dp = getdp(chs1, chs2);
	
	int end = 0;
	int max = 0;
	
	for (int i = 0; i< chs1.length; i++){
		for (int j =0; j < chs2.length; j++){
			if (dp[i][j] > max){
				end = i;
				max = dp[i][j];
			}
			
		}
	}
	return str1.substring(end-max+1, end+1);
	
}

解法二：从斜线最左上的位置想右下依次计算每个位置的值

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42

public String lcat2(String str1, String str2){
	if (str1 == null || str2 == null || str1 == "" || str2 == ""){
		return "";
	}
	char[] chs1 = str1.toCharArray();
	char[] chs2 = str2.toCharArray();
	
	int row = 0;
	int col = chs2.length -1;
	int max = 0;
	int end = 0;
	
	while(row < chs1.length){
		int i = row;
		int j = col;
		int len = 0;
		while (i < chs1.length && j < chs2.length){
			if (chs1[i] != chs2[j]){
				len = 0;
			}else{
				len++;
			}
			
			//记录最大值，以及结束字符的位置
			if (len > max) {
				end = i;
				max = len;
			}
			
			i++;
			j++;
		}
		if (col > 0){
			col--;
		}
		else{
			row++;
		}
	}
	return str1.substring(end -max +1,end +1);
}