https://github.com/tdoly/The-Art-Of-Programming-by-July/blob/master/ebook/zh/31.0.md
https://github.com/julycoding/The-Art-Of-Programming-By-July
Related: Wildcard Matching, Leetcode
https://github.com/julycoding/The-Art-Of-Programming-By-July
Related: Wildcard Matching, Leetcode
字符串匹配问题,给定一串字符串,按照指定规则对其进行匹配,并将匹配的结果保存至output数组中,多个匹配项用空格间隔,最后一个不需要空格。
要求:
- 匹配规则中包含通配符?和*,其中?表示匹配任意一个字符,*表示匹配任意多个(>=0)字符。
- 匹配规则要求匹配最大的字符子串,例如a*d,匹配abbdd而非abbd,即最大匹配子串。
- 匹配后的输入串不再进行匹配,从当前匹配后的字符串重新匹配其他字符串。
请实现函数:
char* my_find(char input[], char rule[])
举例说明:
注意事项:
1. 自行实现函数my_find,勿在my_find函数里夹杂输出,且不准用C、C++库,和Java的String对象;
2. 请注意代码的时间,空间复杂度,及可读性,简洁性;
3. input=aaa,rule=aa时,返回一个结果aa,即可。
1. 自行实现函数my_find,勿在my_find函数里夹杂输出,且不准用C、C++库,和Java的String对象;
2. 请注意代码的时间,空间复杂度,及可读性,简洁性;
3. input=aaa,rule=aa时,返回一个结果aa,即可。
1. 本题与上述第三十章的题不同,上题字符串转换成整数更多考察对思维的全面性和对细节的处理,本题则更多的是编程技巧。闲不多说,直接上代码:
//copyright@cao_peng 2013/4/23
int str_len(char *a) { //字符串长度
if (a == 0) {
return 0;
}
char *t = a;
for (;*t;++t)
;
return (int) (t - a);
}
void str_copy(char *a,const char *b,int len) { //拷贝字符串 a = b
for (;len > 0; --len, ++b,++a) {
*a = *b;
}
*a = 0;
}
char *str_join(char *a,const char *b,int lenb) { //连接字符串 第一个字符串被回收
char *t;
if (a == 0) {
t = (char *) malloc(sizeof(char) * (lenb + 1));
str_copy(t, b, lenb);
return t;
}
else {
int lena = str_len(a);
t = (char *) malloc(sizeof(char) * (lena + lenb + 2));
str_copy(t, a, lena);
*(t + lena) = ' ';
str_copy(t + lena + 1, b, lenb);
free(a);
return t;
}
}
int canMatch(char *input, char *rule) { // 返回最长匹配长度 -1表示不匹配
if (*rule == 0) { //已经到rule尾端
return 0;
}
int r = -1 ,may;
if (*rule == '*') {
r = canMatch(input, rule + 1); // *匹配0个字符
if (*input) {
may = canMatch(input + 1, rule); // *匹配非0个字符
if ((may >= 0) && (++may > r)) {
r = may;
}
}
}
if (*input == 0) { //到尾端
return r;
}
if ((*rule == '?') || (*rule == *input)) {
may = canMatch(input + 1, rule + 1);
if ((may >= 0) && (++may > r)) {
r = may;
}
}
return r;
}
char * my_find(char input[], char rule[]) {
int len = str_len(input);
int *match = (int *) malloc(sizeof(int) * len); //input第i位最多能匹配多少位 匹配不上是-1
int i,max_pos = - 1;
char *output = 0;
for (i = 0; i < len; ++i) {
match[i] = canMatch(input + i, rule);
if ((max_pos < 0) || (match[i] > match[max_pos])) {
max_pos = i;
}
}
if ((max_pos < 0) || (match[max_pos] <= 0)) { //不匹配
output = (char *) malloc(sizeof(char));
*output = 0; // \0
return output;
}
for (i = 0; i < len;) {
if (match[i] == match[max_pos]) { //找到匹配
output = str_join(output, input + i, match[i]);
i += match[i];
}
else {
++i;
}
}
free(match);
return output;
}
2. 本题也可以直接写出DP(Dynamic Programming, 动态规划)方程,如下代码所示:
//copyright@chpeih 2013/4/23
char* my_find(char input[], char rule[])
{
//write your code here
int len1, len2;
for(len1 = 0; input[len1]; len1++);
for(len2 = 0; rule[len2]; len2++);
int MAXN = len1 > len2 ? (len1+1) : (len2+1);
int **dp;
//dp[i][j]表示字符串1和字符串2分别以i j结尾匹配的最大长度
//记录dp[i][j]是由之前那个节点推算过来 i*MAXN+j
dp = new int *[len1+1];
for (int i = 0;i<=len1;i++)
{
dp[i] = new int[len2+1];
}
dp[0][0] = 0;
for(int i = 1; i <= len2; i++)
dp[0][i] = -1;
for(int i = 1; i <= len1; i++)
dp[i][0] = 0;
for (int i = 1; i <= len1; i++)
{
for (int j = 1; j <= len2; j++)
{
if(rule[j-1] == '*'){
dp[i][j] = -1;
if (dp[i-1][j-1] != -1)
{
dp[i][j] = dp[i-1][j-1] + 1;
}
if (dp[i-1][j] != -1 && dp[i][j] < dp[i-1][j] + 1)
{
dp[i][j] = dp[i-1][j] + 1;
}
}else if (rule[j-1] == '?')
{
if(dp[i-1][j-1] != -1){
dp[i][j] = dp[i-1][j-1] + 1;
}else dp[i][j] = -1;
}
else
{
if(dp[i-1][j-1] != -1 && input[i-1] == rule[j-1]){
dp[i][j] = dp[i-1][j-1] + 1;
}else dp[i][j] = -1;
}
}
}
int m = -1;//记录最大字符串长度
int *ans = new int[len1];
int count_ans = 0;//记录答案个数
char *returnans = new char[len1+1];
int count = 0;
for(int i = 1; i <= len1; i++)
if (dp[i][len2] > m){
m = dp[i][len2];
count_ans = 0;
ans[count_ans++] = i-m;
}else if(dp[i][len2] != -1 && dp[i][len2] == m){
ans[count_ans++] = i-m;
}
if (count_ans!=0)
{
int len = ans[0];
for (int i = 0;i < m;i++)
{
printf("%c",input[i+ans[0]]);
returnans[count++] = input[i+ans[0]];
}
for (int j = 1;j<count_ans;j++)
{
printf(" ");
returnans[count++] = ' ';
len = ans[j];
for (int i = 0;i<m;i++)
{
printf("%c",input[i+ans[j]]);
returnans[count++] = input[i+ans[j]];
}
}
printf("\n");
returnans[count++] = '\0';
}
return returnans;
}