2017/11/5 Leetcode 日记

2017/11/5 Leetcode 日记

476. Number Complement

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

Note:

1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
2. You could assume no leading zero bit in the integer’s binary representation.

Solutions:

class Solution {
public:
    int findComplement(int num) {
        unsigned mask = ~0;
        while(mask & num) mask <<=1;
        return ~mask & ~num;
    }
};

class Solution:
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        i = 1
        while(i<=num):
            i<<=1
        return (i-1)^num

557. Reverse Words in a String III

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Solutions:

class Solution {
public:
    string reverseWords(string s) {
        int index = 0, max;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == ' ' || i == s.size()-1){
                max = i-1;
                if (i == s.size()-1) max = i;
                for(int j = index; j < max; j++, max--){
                    char temp = s[j];
                    s[j] = s[max];
                    s[max] = temp;
                }
                index = i+1;
            }
        }
    return s;
    }
};

class Solution:
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        a = ' '
        s = a.join(x[::-1] for x in s.split())
        return s

500. Keyboard Row

Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

(要你判断能用键盘一行打出来的单词,KeyboardMan必备技能啊)

Solutions:

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        unordered_set<char> set1 = {'q', 'w','e', 'r', 't', 'y', 'u', 'i', 'o', 'p'};
        unordered_set<char> set2 = {'s', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'a'};
        unordered_set<char> set3 = {'z', 'x', 'c', 'v', 'b', 'n', 'm'};
        vector<unordered_set<char>> sets = {set1, set2, set3};
        vector<string> nes;
        for(string& word : words){
            int raw = -1;
            for(int i = 0; i<3; i++){
                if(sets[i].count(tolower(word[0])) > 0)
                    raw = i;
            }
            if(raw == -1) continue;//除0，1，2外的第四种可能
            bool valid = true;
            for(int j = 1, sz = word.size(); j < sz; j++){
                if(sets[raw].count(tolower(word[j])) == 0)
                    valid = false;
            }
            if(valid)nes.push_back(word);
        }
        return nes;
    }
};

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        return filter(re.compile('(?i)([qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*)$').match, words)