http://www.cnblogs.com/nailperry/p/4752987.html

 

/**     * @param n     * @param x [1,9]     * @return     */    public int NumberOfXBetween1AndN_Solution(int n,int x) {        if(n<0||x<1||x>9)            return 0;        int high,low,curr,tmp,i = 1;        high = n;        int total = 0;        while(high!=0){            high = n/(int)Math.pow(10, i);// 获取第i位的高位            tmp = n%(int)Math.pow(10, i);            curr = tmp/(int)Math.pow(10, i-1);// 获取第i位            low = tmp%(int)Math.pow(10, i-1);// 获取第i位的低位            if(curr==x){                total+= high*(int)Math.pow(10, i-1)+low+1;            }else if(curr