leetcode-Warm Up Contest-Aug.21

leetcode   地址： https://leetcode.com/contest/detail/1

(1)-- Lexicographical Numbers

Given an integer n, return 1 - n in lexicographical order.

For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].

Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.

好久没有上leetcode了， 突然发现leetcode开始搞竞赛了， 做了 热身赛的第一题，

简单的leetcode风格的题目，使用dfs，深度优先遍历，字典序的经典做法。 

class Solution {
public:
    void dfs(int cur, int n, vector<int> &ret){
        if(cur > n){ return; }
        ret.push_back(cur); 
        for(int i=0; i<=9; i++){
            dfs(cur*10+i, n, ret); 
        }
    }

    vector<int> lexicalOrder(int n) {
        vector<int> t; 
        for(int i=1; i<=9; i++){
            dfs(i, n, t); 
        }
        return t; 
    }
};

(2)  First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

Note: You may assume the string contain only lowercase letters.

查找字符串中的第一个单一字符，

典型的hash 方法， 字符串只含有lowercase的字符， 构造一个26维度的数组，作为hash table。 

算法复杂度为O(n)， n为字符串的长度，就是扫一次字符串就可以得到。 

class Solution {
public:
    int firstUniqChar(string s) {
        int vis[26] = {0}; 
        for(int i=0; i<s.length(); i++){
            if(vis[s[i]-'a'] == 0){
                vis[s[i]-'a'] = i+1; 
            }else{
                vis[s[i]-'a'] = -1; 
            }
        }
        int ans = 1000000; 
        bool flag = false; 
        for(int i=0; i<26; i++){
            if(vis[i] != 0 && vis[i] != -1 && vis[i] < ans){
                ans = vis[i];
                flag = true; 
            }
        }
        if(flag){
            ans = ans - 1; 
        }else{
            ans = -1; 
        }
        return ans; 
    }
};

(3), Longest Absolute File Path

求出最长的文件路径（记得要加上dir与dir之间的 ‘/’ ， 也算是一个字符）， 明显字符串的组成就是dfs的方式，不妨利用dfs数组进行一个遍历。

时间复杂度： O(n)， 扫描字符串数组的长度n。 

class Solution {
public:
    void dfs(int cur, int depth, int *a, int& ans, string& s){
        int isFile, i = cur; 
        while(i < s.length()){
            if(s[i] == '
'){
                isFile = 0; 
                break; 
            }else if(s[i] == '.'){
                isFile = 1; 
                break; 
            }
            i++; 
        }
        if(isFile == 1){
            int tmp = 0; 
            for(int j=0; j<depth; ++j){
                tmp += a[j] + 1; 
            }
            while(i<s.length() && s[i] != '
'){
                i++; 
            }
            tmp += i - cur;
            if( tmp > ans){
                ans = tmp; 
            }
            if(i < s.length() && s[i] == '
'){
                int j = i+1, cnt = 0; 
                while(j<s.length() && s[j] =='	'){
                    j = j + 1;  cnt = cnt + 1; 
                }
                if(j < s.length() ){
                    dfs(j, cnt, a, ans, s); 
                }
            }
        }else{
            a[depth] = i - cur; 
            if(s[i]=='
'){
                int j = i+1, cnt = 0; 
                while(j<s.length() && s[j]=='	'){
                    j = j + 1; 
                    cnt = cnt + 1; 
                }
                if(j < s.length() ){
                    dfs(j, cnt, a, ans, s); 
                }
            }
        }
    }

    int lengthLongestPath(string input) {
        int *a = new int[100]; 
        for(int i=0; i<100; i++){
            a[i] = 0; 
        }
        int ans = 0; 
        dfs(0, 0, a, ans, input); 
        delete[] a; 
        return ans; 
    }
};