字符串處理

字符串處理

---

比較版本號

If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

0.1 < 1.1 < 1.2 < 13.37

# Ruby
# @param {String} version1
# @param {String} version2
# @return {Integer}
def compare_version(version1, version2)
    v1, v2 = version1.split('.'), version2.split('.')
    if (temp = v1.size - v2.size) > 0
        temp.times do v2 << '0' end
    else 
        (-temp).times do v1 << '0' end
    end
    v1.size.times do |i|
        temp = v1[i].to_i - v2[i].to_i
        next if temp == 0
        return temp > 0 ? 1 : -1
    end
    0
end

注意會有1.1 > 1 和1.0 = 1這樣的判例就行了
Compare Version Numbers

---

最長共同前綴

給出一個由多個字符串組成的數組，返回所有字符串的最長共同前綴。

# Ruby
# @param {String[]} strs
# @return {String}
def longest_common_prefix(strs)
    return '' if strs == []
    return strs[0] if strs.length == 1
    sorted = strs.sort{ |a, b| a.length <=> b.length }
    result = ''
    s = sorted[0]
    s.length.times do |i|
        for j in 1...sorted.length
            if sorted[j][0..i] != s[0..i] then
                return result
            end
        end
        result = s[0..i]
    end
    result
end

Longest Common Prefix

---

最後一個單詞的長度

For example,
Given s = “Hello World”,
return 5.
If the last word does not exist, return 0.

# Ruby
# @param {String} s
# @return {Integer}
def length_of_last_word(s)
    return 0 if s == ''
    strs = s.split
    return 0 if strs[-1].nil?
    strs[-1].length
end

Length of Last Word

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判斷迴文

For example,
“A man, a plan, a canal: Panama” or “” is a palindrome.
“race a car” is not a palindrome.

/* C solution using  pointer */
bool isPalindrome(char* s) {
    int len = strlen(s);
    if(!len) return true;
    char *p1 = s, *p2 = s + len - 1;
    while(p1 < p2){
        if(!isalnum(*p1)){p1++;continue;}
        if(!isalnum(*p2)){p2--;continue;}
        if(tolower(*p1++) != tolower(*p2--)) return false;
    }
    return true;
}

Valid Palindrome