[Leetcode] 800. Similar RGB Color 解题报告

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本文介绍了一种高效的算法，用于寻找与给定RGB颜色最相似的短格式颜色。通过计算不同颜色之间的相似度，该算法能快速找到最佳匹配项，并详细解释了实现过程。

题目：

In the following, every capital letter represents some hexadecimal digit from 0 to f.

The red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand. For example, "#15c" is shorthand for the color "#1155cc".

Now, say the similarity between two colors "#ABCDEF" and "#UVWXYZ" is -(AB - UV)^2 - (CD - WX)^2 - (EF - YZ)^2.

Given the color "#ABCDEF", return a 7 character color that is most similar to #ABCDEF, and has a shorthand (that is, it can be represented as some "#XYZ"

Example 1:
Input: color = "#09f166"
Output: "#11ee66"
Explanation:  
The similarity is -(0x09 - 0x11)^2 -(0xf1 - 0xee)^2 - (0x66 - 0x66)^2 = -64 -9 -0 = -73.
This is the highest among any shorthand color.

Note:

color is a string of length 7.
color is a valid RGB color: for i > 0, color[i] is a hexadecimal digit from 0 to f
Any answer which has the same (highest) similarity as the best answer will be accepted.
All inputs and outputs should use lowercase letters, and the output is 7 characters.

思路：

我们发现R,G,B三个分量之间是相互独立的，所以可以各个单独进行计算。对于每个分量，假如对应的输入为s，那么最多有三个候选的最近颜色：string(2, s[0] - 1), string(2, s[0])以及string(2, s[0] + 1)。所以我们分别进行比较，选出距离最小的即可。

值得注意的是，当s[0] == '9'或者s[0] == 'a'的时候，我们需要进行特殊处理，否则在16进制转换的时候会出错。

代码：

class Solution {
public:
    string similarRGB(string color) {
        string ret("#");
        for (int i = 0; i <= 2; ++i) {
            char c = getSimilar(color.substr(2 * i + 1, 2));
            ret += string(2, c);
        }
        return ret;
    }
private:
    char getSimilar(string s) {
        int min_diff = abs(stoi(s, NULL, 16) - stoi(string(2, s[0]), NULL, 16));
        char min_char = s[0], c;
        if (s[0] > '0') {
            c = s[0] == 'a' ? '9' : s[0] - 1;
            int diff = abs(stoi(s, NULL, 16) - stoi(string(2, c), NULL, 16));
            if (diff < min_diff) {
                min_diff = diff;
                min_char = c;
            }
        }
        if (s[0] < 'f') {
            c = s[0] == '9' ? 'a' : s[0] + 1;
            int diff = abs(stoi(s, NULL, 16) - stoi(string(2, c), NULL, 16));
            if (diff < min_diff) {
                min_diff = diff;
                min_char = c;
            }
        }
        return min_char;
    }
};