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0st.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;}
};