No-Zero integer is a positive integer that does not contain any 0
in its decimal representation.
Given an integer n
, return a list of two integers [A, B]
where:
A
andB
are No-Zero integers.A + B = n
The test cases are generated so that there is at least one valid solution. If there are many valid solutions you can return any of them.
Example 1:
Input: n = 2
Output: [1,1]
Explanation: A = 1, B = 1. A + B = n and both A and B do not contain any 0 in their decimal representation.
Example 2:
Input: n = 11
Output: [2,9]
Constraints:
2 <= n <= 10^4
这道题说是要把给定的正整数转化为两个不含零位的正整数之和,不含零位是指数字的个十百千位等都不是零。既然是 Easy 的题目,一般来说不会有太复杂的解法,通常来说暴力搜索的方法就可以。这道题也不例外,既然让返回任意一组解,就可以按遍历 {1, n-1}, {2, n-2}, {3, n-3} 等等的顺序来检验是否符合题意。检验是否含有零位可以放到一个子函数中,方法也非常直接,每次通过对 10 取余来获得最低位,判断其是否为0,然后原数字再自除以 10。这样就可以找到符合题意的组合了,参见代码如下:
解法一:
class Solution {
public:
vector<int> getNoZeroIntegers(int n) {
for (int i = 1; i < n; ++i) {
if (!hasZero(i) && !hasZero(n - i)) {
return {i, n - i};
}
}
return {-1, -1};
}
bool hasZero(int n) {
while (n > 0) {
if (n % 10 == 0) return true;
n /= 10;
}
return false;
}
};
再来看一种非暴力搜索的解法,这种方法有点巧妙,貌似用到了些数学方面的知识。这里还是每次取出最低位上的数字,假设为d,同时n自除以 10,若最低位d是0或者1的话,且此时n不是0的话,那么要从高位取个1,变成 10 或者 11 来拆,分别拆成 {1, 9} 或者 {2, 9},同时n要自减1。对于其他的情况,拆成 {1, d-1},注意为了跟原数字保持相同的为,这里拆成的数都要乘以 step,这里的 step 就是此时n缩小的倍数,参见代码如下:
解法二:
class Solution {
public:
vector<int> getNoZeroIntegers(int n) {
int a = 0, b = 0, step = 1;
while (n > 0) {
int d = n % 10;
n /= 10;
if ((d == 0 || d == 1) && n > 0) {
a += step * (1 + d);
b += step * 9;
--n;
} else {
a += step * 1;
b += step * (d - 1);
}
step *= 10;
}
return {a, b};
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1317
参考资料:
https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/
LeetCode All in One 题目讲解汇总(持续更新中...)
标签:1317,Integers,return,10,int,整数,step,integer,integers From: https://www.cnblogs.com/grandyang/p/16757417.html