参考
http://blog.csdn.net/u012102306/article/details/53184446
http://blog.csdn.net/hrn1216/article/details/51534607
最长公共子序列LCS
动态规划状态转移方程式
这里写图片描述
Python
递归
def LCS(a, b):
if a == '' or b == '':
return ''
elif a[-1] == b[-1]:
return LCS(a[:-1], b[:-1]) + a[-1]
else:
sol_a = LCS(a[:-1], b)
sol_b = LCS(a, b[:-1])
if len(sol_a) > len(sol_b):
return sol_a
return sol_b
if __name__ == "__main__":
a = 'abc'
print(a[::-1])
print(LCS(a,a[::-1]))
动态规划
def lcs_dp(self, input_x, input_y):
# input_y as column, input_x as row
dp = [([0] * (len(input_y)+1)) for i in range(len(input_x)+1)]
for i in range(1, len(input_x)+1):
for j in range(1, len(input_y)+1):
if i == 0 or j == 0: # 在边界上，自行+1
dp[i][j] = 1
elif input_x[i-1] == input_y[j-1]: # 不在边界上，相等就加一
dp[i][j] = dp[i - 1][j - 1] + 1
else: # 不相等
dp[i][j] = max(dp[i - 1][j], dp[i][j -1])
for dp_line in dp:
print(dp_line)
return dp[-1][-1]
Java
动态规划
public static int lcs(String str1, String str2) {
int len1 = str1.length();
int len2 = str2.length();
int c[][] = new int[len1+1][len2+1];
for (int i = 0; i <= len1; i++) {
for( int j = 0; j <= len2; j++) {
if(i == 0 || j == 0) {
c[i][j] = 0;
} else if (str1.charAt(i-1) == str2.charAt(j-1)) {
c[i][j] = c[i-1][j-1] + 1;
} else {
c[i][j] = max(c[i - 1][j], c[i][j - 1]);
}
}
}
return c[len1][len2];
}
最长回文子串
动态规划状态转移方程式
这里写图片描述
Python
动态规划
同时输出长度和字符串
class LCS3:
def lcs3_dp(self, input_x, input_y):
# input_y as column, input_x as row
dp = [([0] * (len(input_y)+1)) for i in range(len(input_x)+1)]
maxlen = maxindex = 0
for i in range(1, len(input_x)+1):
for j in range(1, len(input_y)+1):
if i == 0 or j == 0: # 在边界上，自行+1
dp[i][j] = 0
if input_x[i-1] == input_y[j-1]:
dp[i][j] = dp[i - 1][j - 1] + 1
if dp[i][j] > maxlen: # 随时更新最长长度和长度开始的位置
maxlen = dp[i][j]
maxindex = i - maxlen
# print('最长公共子串的长度是:%s' % maxlen)
# print('最长公共子串是:%s' % input_x[maxindex:maxindex + maxlen])
else:
dp[i][j] = 0
for dp_line in dp:
print(dp_line)
return maxlen, input_x[maxindex:maxindex + maxlen]
if __name__ == '__main__':
lcs3 = LCS3()
print(lcs3.lcs_dp('cabdec','cbdec'))
运行结果
[1, 0, 0, 0, 1]
[0, 0, 0, 0, 0]
[0, 1, 0, 0, 0]
[0, 0, 2, 0, 0]
[0, 0, 0, 3, 0]
[1, 0, 0, 0, 4]
bdec
Java
动态规划(懒得加上返回字符串了)
public static int lcs3(String str1, String str2) {
int len1 = str1.length();
int len2 = str2.length();
int result = 0; //记录最长公共子串长度
int c[][] = new int[len1+1][len2+1];
for (int i = 0; i <= len1; i++) {
for( int j = 0; j <= len2; j++) {
if(i == 0 || j == 0) {
c[i][j] = 0;
} else if (str1.charAt(i-1) == str2.charAt(j-1)) {
c[i][j] = c[i-1][j-1] + 1;
result = max(c[i][j], result);
} else {
c[i][j] = 0;
}
}
}
return result;
}
