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Java for LeetCode 233 Number of Digit One

时间:2015-10-20 21:02:04      阅读:180      评论:0      收藏:0      [点我收藏+]

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Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

解题思路:

递归

static public int countDigitOne(int n) {
		if (n == 0)
			return 0;
		if (n < 10)
			return 1;
		int length = 0;
		int firstNum = n;
		while (firstNum >= 10) {
			firstNum /= 10;
			length++;
		}
		int basic = (int) Math.pow(10, length);
		if (firstNum > 1) {
			int tmp1 = countDigitOne(basic - 1);
			int tmp2 = basic;
			int tmp3 = countDigitOne(n - basic * firstNum);
			return firstNum * tmp1 + tmp2 + tmp3;
		} else {
			int tmp1 = countDigitOne(basic - 1);
			int tmp2 = n + 1 - basic;
			int tmp3 = countDigitOne(n - basic);
			return tmp1 + tmp2 + tmp3;
		}

	}

 

Java for LeetCode 233 Number of Digit One

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原文地址:http://www.cnblogs.com/tonyluis/p/4895855.html

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