algo-aha

-2进制转换

• https://leetcode.com/problems/convert-to-base-2/

string baseNeg2(int N) {
    // 不能用%-2操作，因为需要余数为正
    string ans;
    while (N) {
        int rem = N & 1;
        ans = to_string(rem) + ans;
        N = -(N >> 1);
    }
    return !ans.empty() ? ans : "0";
}

删除9

• https://leetcode.com/problems/remove-9/

Start from integer 1, remove any integer that contains 9 such as 9, 19, 29...

So now, you will have a new integer sequence: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, ...

Given a positive integern, you need to return the n-th integer after removing. Note that 1 will be the first integer.

Example 1:

Input:
 9

Output:
 10

Hint: n will not exceed9 x 10^8.

int newInteger(int n) {
    // 删除9后的数正好就是九进制数，这题就是把十进制转成九进制
    // 扩展：如果删除7，还是九进制，只是结果需映射7=>8,8=>9。
    int ans = 0, base = 1;
    while (n) {
        ans += (n % 9) * base; // 从后往前拼上
        n /= 9;
        base *= 10;
    }
    return ans;
}

相似的RGB颜色

• https://leetcode.com/problems/similar-rgb-color/

string similarRGB(string color) {
    return "#" + similarColor(color.substr(1, 2))
        + similarColor(color.substr(3, 2))
        + similarColor(color.substr(5, 2));
}

string similarColor(const string &ab) {
    const string mapping = "0123456789abcdef";
    // #ab要变成#xx，#xx=16*x+x=17x，所以要找最接近#ab的17的倍数
    int num = stoi(ab, NULL, 16);
    int x = (num + 8) / 17; // "四舍五入"
    return string(2, mapping[x]);
}