题解 | #字符串排序#
字符串排序
https://www.nowcoder.com/practice/5190a1db6f4f4ddb92fd9c365c944584
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
//思路：先将小写字母全部化为大写，并对变化的元素设置标记，再进行稳定的排序，对于非字母直接跳过排序
//最后根据标记将变化的大写字母还原为小写字母
char a[1000];
char change[1000] = {0};
int main() {
gets(a);
int n = strlen(a);
for (int i = 0; i < n; i++) { //小写化为大写
if (a[i] >= 97 && a[i] <= 122) {
a[i] = a[i] - 32;
change[i] = 1;
}
}
for (int i = 0; i < n - 1; i++) {
for (int j = n - 1; j > i; j--) {
int tmp = j - 1;
if (a[j] < 65 || a[j] > 90) {
continue;
}
while (a[tmp] < 65 || a[tmp] > 90 &&
tmp >= 0) { //注意：必须加tmp>=0的判断，不然就数组越界了，也就是oj报错的原因
tmp--;
}
if (tmp < 0) {
continue;
}
if (a[tmp] > a[j]) {
int tmp2 = change[tmp];
char tmp1 = a[tmp];
change[tmp] = change[j];
change[j] = tmp2;
a[tmp] = a[j];
a[j] = tmp1;
}
}
}
for (int i = 0; i < n; i++) {
if (change[i] == 1) {
a[i] = a[i] + 32;
}
}
printf("%s", a);
}
查看23道真题和解析
