[LeetCode] 800. Similar RGB Color 相似的红绿蓝颜色
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.
大写字母组成的16进制字符串表示RGB颜色，可以简写为'#XYZ'形式的，两个颜色的相似度由给的公式计算得出。给一个颜色，求和它最相似的颜色，返回的是7个字符表示的颜色。
解法1：暴力brute force.
解法2：取余取模
解法3： 将字符串分为3个部分，对每个部分找个最相近的 ‘XX’ 格式的值即可。对于每一个 ‘XY’ 值，最近的一定在 ‘XX’, ‘(X-1)(X-1)’, ‘(X+1)(X+1)’ 之中，所以计算出这三个值取最近的即可。
Java: 暴力
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(16) for j in range(16) for k in range(16)]
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>d:
min=d
res=s
return '#'+''.join(res)
Python: 暴力
class Solution(object):
def similarRGB(self, color):
"""
:type color: str
:rtype: str
"""
ir, ig, ib = (int(color[x: x+2], 16)
for x in (1, 3, 5))
ans = ()
delta = 0x7FFFFFFF
for r in range(16):
for g in range(16):
for b in range(16):
ndelta = sum((ic - c * 17) ** 2
for ic, 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: 解法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, 16), 17)
if r > 8: q += 1
return '{:02x}'.format(17*q)
return '#' + \
rounding(color[1:3]) + \
rounding(color[3:5]) + \
rounding(color[5:7])
Python: 解法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)*2 if c<=9 else chr(c-10+ord('a'))*2
return ret
C++: 暴力，T: O(3 * 16) 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]) * 16 + 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 < 16; ++j) {
int diff = abs(j * 16 + 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, 16);
int idx = num / 17 + (num % 17 > 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, 16);
int idx = num / 17 + (num % 17 > 8 ? 1 : 0);
color[i] = color[i + 1] = (idx > 9) ? (idx - 10 + 'a') : (idx + '0');
}
return color;
}
};
