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:
coloris a string of length7.coloris a valid RGB color: fori > 0,color[i]is a hexadecimal digit from0tof- 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.
Solution #1
class Solution {
private static int[] preset = {0,17,34,51,68,85,102,119,136,153,170,187,204,221,238,255};
private static String[] hex = {"00","11","22","33","44","55","66","77","88","99","aa","bb","cc","dd","ee","ff"};
public String similarRGB(String color) {
String s = "#";
for (int i = 1; i < color.length(); i += 2) {
int min = 0;
int cand = toInt(color.charAt(i)) * 16 + toInt(color.charAt(i + 1));
for (int j = 0; j < preset.length; j++) {
if (Math.abs(cand - preset[j]) < Math.abs(cand - preset[min])) {
min = j;
}
}
s += hex[min];
}
return s;
}
private int toInt(char c) {
int rt = 0;
if (Character.isDigit(c)) {
rt = c - '0';
}else {
rt = 10 + (c - 'a');
}
return rt;
}
}
Solution #2: https://leetcode.com/problems/similar-rgb-color/discuss/119844/Concise-java-solution
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