这道题我一开始的想法是,把数字按位数分开,比如19872分成10000和10001-19872,然后对于10的幂次方的值打表,找规律,但后来发现,这样分开,涉及到的很多琐碎的加一要考虑(就是10001-19872和10000-19872的区别,每一次这样分开,都要加一,就会加多,只加一次,一些数据又不需要加一,很难办),最后看了答案,答案直接按位找规律计算,不用递归分开,省事很多。(https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/mian-shi-ti-43-1n-zheng-shu-zhong-1-chu-xian-de-2/)
答案讲解的很清晰,不多写了
public int countDigitOne(int n) {
int high = n/10,cur = n%10,low = 0;
int digit = 1;
int res=0;
while(high!=0||cur!=0){
if(cur == 0){
res += high*digit;
}else if(cur == 1){
res+= (high*digit + low+1);
}else{
res += high*digit + digit;
}
low +=(cur*digit);
cur = high%10;
high/=10;
digit*=10;
}
return res;
}