1~n整数中1出现的次数（Java实现）

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本文介绍了一种高效的算法，用于计算从1到任意整数N范围内所有数字中数字1出现的总次数。通过优化的解决方案，算法的时间复杂度降低至O(logN)，显著提高了大规模数据处理的效率。


public class E43NumberOf1Between1AndN {
    //1-N整数中含1的个数

    public static int numberOf1(int n){ //时间复杂度为NO(log(N))
        if (n <= 0)
            return -1;
        int count = 0;
        for (int i = 1; i <= n; i ++){
            if (contains1(i) != 0)
                count += contains1(i);
        }
        return count;
    }

    private static int contains1(int number){
        int count = 0;
        while(number != 0){
            if (number % 10 == 1)
                count ++;
            number /= 10;
        }
        return count;
    }

    public static int numberOf1_BetterSollution(int n){ //时间复杂度为O(logN)(底为10)
        //将每一个位次上（个位+十位+...）1出现的次数相加
        if (n < 1)
            return 0;
        int count = 0;
        int base = 1;
        int round = n;
        int weight = round % 10;
        while(round > 0){
            round /= 10;
            count += round * base;
            if (weight == 1)
                count += 1 + n % 10;
            else if (weight > 1)
                count += base;
            base *= 10;
            weight = round % 10;
        }
        return count;
    }

    //测试用例
    public static void main(String[] args){
        System.out.println(E43NumberOf1Between1AndN.numberOf1(12)); //5
        System.out.println(E43NumberOf1Between1AndN.numberOf1_BetterSollution(12)); //5
        System.out.println(E43NumberOf1Between1AndN.numberOf1_BetterSollution(21345)); //18481
    }
}