标签：cte 括号匹配 target pop ems href des desc view
Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
自己代码：
class Solution { public: bool isValid(string s) { stack<char> parent; for(char c:s){ if(c ==‘[‘||c == ‘(‘ || c == ‘{‘) parent.push(c); //左括号进栈 else if(parent.empty()) return false; else{ //有右括号且与栈顶匹配，则栈顶元素出栈 if(c ==‘]‘ && parent.top() == ‘[‘) parent.pop(); else if(c ==‘}‘ && parent.top() == ‘{‘) parent.pop(); else if(c ==‘)‘ && parent.top() == ‘(‘) parent.pop(); else return false; } } if(parent.empty()) return true; else return false; } };
巧妙的方法，discussion区发现：
遇到左括号，则使对应右括号进栈，例如：遇到“{”，进栈“}”。遇到“[”，进栈“]”。遇到“(”，进栈“)”。
非右括号，则看它是否与栈顶元素相等，相等即匹配。
public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); for (char c : s.toCharArray()) { if (c == ‘(‘) //遇到左右括号，使括号进栈 stack.push(‘)‘); else if (c == ‘{‘) stack.push(‘}‘); else if (c == ‘[‘) stack.push(‘]‘); else if (stack.isEmpty() || stack.pop() != c) //遇到非右括号，若栈空，返回false。否则弹出栈顶元素，看其是否与当前元素相等，否则错误 return false; } return stack.isEmpty(); }
用switch语句写：
bool isValid(string s) { stack<char> paren; for (char& c : s) { switch (c) { case ‘(‘: case ‘{‘: case ‘[‘: paren.push(c); break; case ‘)‘: if (paren.empty() || paren.top()!=‘(‘) return false; else paren.pop(); break; case ‘}‘: if (paren.empty() || paren.top()!=‘{‘) return false; else paren.pop(); break; case ‘]‘: if (paren.empty() || paren.top()!=‘[‘) return false; else paren.pop(); break; default: ; // pass } } return paren.empty() ; }
标签：cte 括号匹配 target pop ems href des desc view
原文地址：http://www.cnblogs.com/hozhangel/p/7822723.html
