力扣793. 阶乘函数后 K 个零

Problem: 793. 阶乘函数后 K 个零

题目描述

思路即解法

1.根据题意可知即是要求取满足条件的n最小是多少，最大是多少，最大值和最小值一减，就可以算出来有多少个n满足条件了。
2.由于题目中的阶乘存在单调性则可以使用二分查找来解决，具体的利用一个辅助函数求取出每一个n的阶乘后的0的个数，再利用二分查找找出满足条件的n的左边界和有边界

复杂度

时间复杂度:

$O(1)$；若不限制数据范围则为$O(logN\times logN)$

空间复杂度:

$O(1)$

Code

class Solution {
    /**
     * Preimage Size of Factorial Zeroes Function
     *
     * @param k Given number
     * @return int
     */
    public int preimageSizeFZF(int k) {
        // The difference between the left and right boundaries + 1 is the answer
        return (int) (right_bound(k) - left_bound(k) + 1);
    }

    /**
     * Search for the left boundary of trailingZeroes(n) == K
     *
     * @param target The target
     * @return long
     */
    private long left_bound(int target) {
        long low = 0;
        long high = Long.MAX_VALUE;
        while (low < high) {
            long mid = low + (high - low) / 2;
            if (trailingZeroes(mid) < target) {
                low = mid + 1;
            } else if (trailingZeroes(mid) > target) {
                high = mid;
            } else {
                high = mid;
            }
        }
        return low;
    }

    /**
     * Search for the right boundary of trailingZeroes(n) == K
     *
     * @param target The target
     * @return long
     */
    private long right_bound(int target) {
        long low = 0;
        long high = Long.MAX_VALUE;
        while (low < high) {
            long mid = low + (high - low) / 2;
            if (trailingZeroes(mid) < target) {
                low = mid + 1;
            } else if (trailingZeroes(mid) > target) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low - 1;
    }

    /**
     * Find the number of zeros after the factorial of a given number
     *
     * @param n Given number
     * @return long
     */
    private long trailingZeroes(long n) {
        long res = 0;
        for (long d = n; d / 5 > 0; d = d / 5) {
            res += d / 5;
        }
        return res;
    }
}