NYOJ915 +-字符串

原题链接

主要思路：从左到右，逐个比较，若有不同，标记此不同地点，并向右搜寻首个相同点，从该点开始挨个与左边位置交换并统计交换次数。

#include <stdio.h>
#include <string.h>
#define MAX 5000 + 2
char str[MAX], str2[MAX];

int find(int i){
int j = i, count = 0;
char t;
while(str[j] != str2[i])
++j;
while(j > i){
t = str[j];
str[j] = str[j - 1];
str[j - 1] = t;
--j;
++count;
}
return count;
}

int main(){
int count;
while(scanf("%s%s", str, str2) == 2){
int a = 0, b = 0;
for(int i = 0; str[i] != '\0'; ++i)
if(str[i] == '-') ++a;
for(int i = 0; str2[i] != '\0'; ++i)
if(str2[i] == '-') ++b;
if(a != b){
printf("-1\n");
continue;
}
count = 0;
for(int i = 0; str2[i] != '\0'; ++i){
if(str[i] != str2[i])
count += find(i);
}
printf("%d\n", count);
}
return 0;
}

优化后的代码：思路有所改变。不再模拟整个过程，时间大幅缩短。

#include <stdio.h>
#include <string.h>
#define MAX 5000 + 2
char str[MAX], str2[MAX];

int main(){
int count, ok, i, j;
while(scanf("%s%s", str, str2) == 2){
count = 0;
for(i = 0, ok = 1; str2[i] != '\0'; ++i){
if(str[i] != str2[i]){
for(j = i + 1, ok = 0; str[j] != '\0'; ++j)
if(str[j] == str2[i]){
count += j - i;
str[j] = str[i];
ok = 1;
break;
}
}
if(!ok){
printf("-1\n");
break;
}
}
if(ok) printf("%d\n", count);
}
return 0;
}