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leetcode部分解题思路

admin 11月 13, 2020 0

1. A+B problem

Write a function that add two numbers A and B. You should not use + or any arithmetic operators.

加法分两部分：按位异或得到当前位的值，按位与得到当前位产生的进位。

int aplusb(int a, int b) {
    // write your code here, try to do it without arithmetic operators.
    return b == 0 ? a : aplusb(a^b, ((a&b) << 1));
}

3. Digit Counts

Count the number of k’s between 0 and n. k can be 0 - 9.
把数字转换成字符串来做，想法很新，值得注意，但是感觉复杂度还是很高。

int digitCounts(int k, int n) {
    // write your code here
    char k_ch=k+'0';
    int count=0;
    for(int i=0;i<=n;++i) {
        string n_str=to_string(i);
        int c=0;
        for(int j=0;j<n_str.size();++j) {
            if(n_str[j]==k_ch) ++c;
        }
        count+=c;
    }
    return count;
}

4. Ugly Number II

Ugly number is a number that only have factors 2, 3 and 5.

这个题还是没理解。下面是别人的解法：http://www.cnblogs.com/grandyang/p/4743837.html

int nthUglyNumber(int n) {
    vector<int> res(1, 1);
    int i2 = 0, i3 = 0, i5 = 0;
    while (res.size() < n) {
        int m2 = res[i2] * 2, m3 = res[i3] * 3, m5 = res[i5] * 5;
        int mn = min(m2, min(m3, m5));
        if (mn == m2) ++i2;
        if (mn == m3) ++i3;
        if (mn == m5) ++i5;
        res.push_back(mn);
    }
    return res.back();
}