leetcode ----Trie/stack专题

Posted 2020-09-28 brucemengbm

一：Implement Trie (Prefix Tree)

题目：

Implement a trie with insert, search, and startsWith methods.

Note:
You may assume that all inputs are consist of lowercase letters a-z.

分析：此题是典型的trie树。能够參见：http://blog.csdn.net/lu597203933/article/details/44227431

代码：

class TrieNode {
public:
    // Initialize your data structure here.
    TrieNode() {
        for(int i = 0; i < 26; i++)
            next[i] = NULL;
        isString = false;
    }
    TrieNode *next[26];
    bool isString;
};

class Trie {
public:
    Trie() {
        root = new TrieNode();
        
    }

    // Inserts a word into the trie.
    void insert(string s) {
        TrieNode *p = root;
        for(int i = 0; i < s.size(); i++){
            if(p->next[s[i]-'a'] == NULL){
                p->next[s[i]-'a'] = new TrieNode();
            }
            p = p->next[s[i]-'a'];
        }
        p->isString = true;
    }

    // Returns if the word is in the trie.
    bool search(string key) {
        TrieNode *p = root;
        for(int i = 0; i < key.size(); i++){
            if(p == NULL) return false;
            p = p->next[key[i]-'a'];
        }
        if(p == NULL || p->isString == false) return false;
        return true;
        
    }

    // Returns if there is any word in the trie
    // that starts with the given prefix.
    bool startsWith(string prefix) {
        TrieNode *p = root;
        for(int i = 0; i <= prefix.size(); i++){
            if(p == NULL) return false;
            p = p->next[prefix[i]-'a'];
        }
        return true;
    }

private:
    TrieNode* root;
};

// Your Trie object will be instantiated and called as such:
// Trie trie;
// trie.insert("somestring");
// trie.search("key");

二：Basic Calculator

题目：

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

分析：这道题能够将括号中面的看做是负负得正，这样用sign记录当前数字前面得符号。是+为1，是-为-1，。然后还要它所在的括号深度的符号，用stack来标记。

同一时候num = num *10 + c - ‘0‘。， 也给出了正向计算一个字符串比方“12342”的数值方法。。。

class Solution {
public:
    int calculate(string s) {
		int sign = 1;         // 当前元素前是+还是-
		stack<char> st;       // 主要是为了考虑括号的深度， 括号前面是+ 则为1否则为0
		st.push(1); 
		int ans = 0;
		int tmp = 0;
		for(int i = 0; i < s.size(); i++){
			char c = s[i];
			if(isdigit(c)){    // 假设是数字  则保存起来
				tmp = tmp * 10 + s[i] - '0'; 
			}
			else if(c == '-' || c == '+'){  
				ans += tmp * sign * st.top();   // 由当前符号和括号外面的符号两者决定！
				sign = (c=='+' ? 1 : -1);
				tmp = 0;
			}else if(c == '('){
				st.push(sign*st.top());   // 当前括号内元素的符号由其前面的各个外层括号符号决定
				sign = 1;   // 括号后面首个是+
			}else if(c ==')'){
				ans += tmp *sign * st.top();
				st.pop();
				tmp = 0;
				sign = 1;
			}
		}
		ans += tmp * sign * st.top();
        return ans;
        
    }
};