采用《编程之美》的算法,但是这个算法会在自动统计置零位。例如9会当作09处理,从而多统计一个0。所以在统计完成后对count的值再调整。
代码如下。


class Solution {
public:
    /*
     * @param : An integer
     * @param : An integer
     * @return: An integer denote the count of digit k in 1..n
     */
    int digitCounts(int k, int n) {
        // write your code here
        int count = 0;
        int base = 1;
        while(n/base>0){
            int cur = (n/base)%10;
            int high = n/(base*10);
            int low = n - (n/base)*base;
            
            if (cur == k){
                count += high * base + low + 1;
            } else if (cur<k){
                    count += high*base;
                }else {
                    count += (high+1)*base;
                }
            base *= 10;
        }
            if (k==0 & n!=0){
                count -= base/10;
            }
            else if (k==0 && n==0){ 
                 count = 1;
            }
        return count;
    }
};

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