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Add C++ solution for leetcode 233.

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	`1`	`+#include<vector>`
	`2`	`+#include<algorithm>`
	`3`	`+#include<iostream>`
	`4`	`+using namespace std;`
	`5`	`+`
	`6`	`+class Solution {`
	`7`	`+public:`
	`8`	`+ int countDigitOne(int n) {`
	`9`	`+ if (n == 0) return 0; /* 0 <= n <= 10^9 */`
	`10`	`+`
	`11`	`+ int curDigit;`
	`12`	`+ int count = 0; // 记录1的数量`
	`13`	`+ long highDigt, lastDigit, weigh = 1; // weigh用来记录置1的位置处于个位、十位、百位、千位等...`
	`14`	`+ while (weigh <= n) // weigh每次乘10相对于从低位向高位方向逐位移动,直到超过n时结束循环`
	`15`	`+ {`
	`16`	`+ highDigt = n / (weigh * 10); // 得到某一个位数置1后的商,即当前数之前的部分`
	`17`	`+ lastDigit = n % weigh; // 得到某一位数置1后的余数,即当前数之后的部分`
	`18`	`+ curDigit = (n / weigh) % 10; // 当前置1的位置原来(置1之前)的数`
	`19`	`+`
	`20`	`+ if (curDigit == 0) // 当前位置的数字=0时`
	`21`	`+ count += highDigt * weigh;`
	`22`	`+ if (curDigit == 1) // 当前位置的数字=1时`
	`23`	`+ count += highDigt * weigh + lastDigit + 1;`
	`24`	`+ if (curDigit > 1) // 当前位置的数字>1时`
	`25`	`+ count += (highDigt + 1) * weigh;`
	`26`	`+`
	`27`	`+ weigh *= 10;`
	`28`	`+ }`
	`29`	`+ return count;`
	`30`	`+ }`
	`31`	`+};`
	`32`	`+`
	`33`	`+// Test`
	`34`	`+int main()`
	`35`	`+{`
	`36`	`+ Solution sol;`
	`37`	`+ int n = 13;`
	`38`	`+ auto res = sol.countDigitOne(n);`
	`39`	`+ cout << res << endl;`
	`40`	`+`
	`41`	`+ return 0;`
	`42`	`+}`

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