[Leetcode] 800. Similar RGB color similar to red-green-blue

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

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

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

Given the color "#ABCDEF" , return a 7 character color that's most similar #ABCDEF to, and have a shorthand (that's, it can be r Epresented 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 length 7 .
• coloris a valid RGB color:for i > 0 , was color[i] a hexadecimal digit from 0 tof
• Any answer which have the same (highest) similarity as the best answer would be accepted.
• All inputs and outputs should use lowercase letters, and the output is 7 characters.

A 16 binary string consisting of uppercase letters represents the RGB color, which can be abbreviated as ' #XYZ ', and the similarity of two colors is calculated by the formula given. Give a color, sum it the most similar color, and return the color represented by 7 characters.

Solution 1: Violence brute force.

Solution 2:

Solution 3: Divide the string into 3 parts and find the nearest ' XX ' format value for each section. For each ' XY ' value, the nearest must be in ' XX ', ' (X-1) (X-1) ', ' (x+1) ' (x+1) ', so calculate these three values to take the nearest one.

Java: Violence

Class Solution:    def similarrgb (self, color): "" "        : Type color:str        : Rtype:str        " ""        r,g,b = Int ( color[1:3],16), int (color[3:5],16), int (color[5:7],16)        a = [' 00 ', ' 11 ', ' 22 ', ' 33 ', ' 44 ', ' 55 ', ' 66 ', ' 77 ', ' 88 ', ' 99 ', ' AA ', ' BB ', ' cc ', ' dd ', ' ee ', ' ff ']        p = [(A[i],a[j],a[k]) for I in range (+) for J in range (+) for K in range (+)]        res , Min = ", 9999999 for        s in P:            d = (int (s[0],16)-R) **2 + (int (s[1],16)-G) **2 + (int (s[2],16)-B) **2            if MIN&G T;d:                min=d                res=s        return ' # ' + '. Join (RES)    

Python: Violence

Class solution (Object):    def similarrgb (self, color): "" "        : Type color:str        : Rtype:str        "        "" IR , IG, IB = (int (color[x:x+2], +) for                      x in (1, 3, 5))        ans = ()        delta = 0x7FFFFFFF for        R in range: 
   for g in range (+):                for B in range (+):                    Ndelta = SUM ((IC-C *) * * 2                                 for ICS, C in Zip (IR, IG, IB), (R , G, B)))                    if Ndelta < delta:                        delta = ndelta                        ans = r, G, b        return ' # ' + '. Join (Hex (c) [2] * 2 for C In ans)

Python: Solution 2 Time:o (1), Space:o (1)

Class solution (Object):    def similarrgb (self, color): "" "        : Type color:str        : Rtype:str        "        "" Def rounding (color):            q, r = divmod (int (color, +), +)            if r > 8:q + = 1            return ' {: 02x} '. Format (17*q)        re Turn ' # ' +                 rounding (Color[1:3]) +                 rounding (Color[3:5]) +                 rounding (Color[5:7])

Python: Solution 3

def similarrgb (self, color):    ret = ' # ' for    I in range (1, 6, 2):        c1, c2 = [Int (_) if ' 0 ' <=_<= ' 9 ' else 10+ord (_)-ord (' a ') for _ in color[i:i+2]]        c = c1+sorted (Enumerate ([ABS ((C1*16+C2)-(x*16+x)) for x in [C1-1, C1, c1+1] ]), Key=lambda _:_[1]) [0][0]-1        ret + = str (c) * If c<=9 else Chr (C-10+ord (' a ')) * *    return RET

C + +: Violence, T:o (3 *) S:o (1)

Class Solution {public:  string Similarrgb (string color) {    const string hex{"0123456789abcdef"};    Vector<int> RGB (3, 0);    for (int i = 0; i < 3; ++i)      rgb[i] = hex.find (color[2 * i + 1]) * + hex.find (color[2 * i + 2]);        String ans (7, ' # ');        for (int i = 0; i < 3; ++i) {      int. best = Int_max;      for (int j = 0; J < ++j) {        int diff = ABS (J * + j-rgb[i]);        if (diff >= Best) continue;        Best = diff;        ans[2 * i + 1] = ans[2 * i + 2] = Hex[j];      }    }    return ans;  };

C++:

Class Solution {public:    string Similarrgb (string color) {        return "#" + Helper (Color.substr (1, 2)) + helper (color. SUBSTR (3, 2) + Helper (Color.substr (5, 2));    }    String Helper (String str) {        string dict = "0123456789abcdef";        int num = Stoi (str, nullptr, +);        int idx = NUM/17 + (num% > 8? 1:0);        return string (2, Dict[idx]);}    ;

C++:

Class Solution {public:    string Similarrgb (string color) {for        (int i = 1; i < color.size (); i + = 2) {            int num = Stoi (Color.substr (i, 2), nullptr, +);            int idx = NUM/17 + (num% > 8? 1:0);            Color[i] = color[i + 1] = (idx > 9)? (idx-10 + ' A '): (idx + ' 0 ');        }        return color;    }};

　　

　

[Leetcode] 800. Similar RGB color similar to red-green-blue