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nonRepeat.cpp
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80 lines (74 loc) · 1.48 KB
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#include <cstdio>
#include <string>
using namespace std;
// Naive
void getNonRepeat1(int num) {
bool flag;
while(1) {
flag = true;
num++;
int pre = num % 10, temp = num / 10;
while(temp) {
int cur = temp % 10;
temp = temp / 10;
if (cur == pre) {
flag = false;
break;
}
pre = cur;
}
if (flag) break;
}
printf("%d\n", num);
}
// Naive
void getNonRepeat2(int num) {
string s = '0' + to_string(num);
int i = 1, len = s.size();
bool first = true;
while (1) {
while (i + 1 < len && s[i] != s[i + 1]) i++;
// 原数本身就是一个不重复数,则在其最低位+1
//+1时进行判断,如果个位为9,则变为0,再进位,否则直接+1
//+1后判断会不会跟前一位重复
if (i == len - 1 && first) {
if (s[i] == '9') {
s[i] = '0';
i--;
}
s[i] += 1;
i--;
first = false;
continue;
}
// 不是首次扫描到此,则直接退出,表示当前数为最小不重复数
else if (i == len - 1) break;
// 重复的是'99',此两位置0,高位直接+1
if (s[i + 1] == '9') {
s[i] = s[i + 1] = '0';
s[i - 1] += 1;
i -= 2;
}
else {
s[i + 1] += 1;
i += 2;
bool flag = true;
while (i < len) {
if (flag) s[i++] = '0';
else s[i++] = '1';
flag = !flag;
}
break;
}
}
s[0] == '0'? printf("%s\n", s.substr(1).c_str()) : printf("%s\n", s.c_str());
}
int main()
{
int num;
while (scanf("%d", &num) != EOF) {
getNonRepeat1(num);
getNonRepeat2(num);
}
return 0;
}