20. Valid Parentheses
Given a string containing just the characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: “()”
Output: true
Example 2:
Input: “()[]{}”
Output: true
Example 3:
Input: “(]”
Output: false
Example 4:
Input: “([)]”
Output: false
Example 5:
Input: “{[]}”
Output: true
C++
class Solution {
public:
bool isValid(string s) {
stack<char> r;
for(int i=0;i<s.length();i++)
{
if(s[i]=='(' || s[i]=='[' || s[i]=='{')
r.push(s[i]);
else
{
if(r.empty())
return false;
char t=r.top();
r.pop();
if(s[i]==')'&&t!='(') return false;
if(s[i]==']'&&t!='[') return false;
if(s[i]=='}'&&t!='{') return false;
}
}
if(!r.empty()) return false;
return true;
}
};
155. Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) – Push element x onto stack.
pop() – Removes the element on top of the stack.
top() – Get the top element.
getMin() – Retrieve the minimum element in the stack.
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
设计一个最小栈，实现push, pop, top, getMin四个功能。相比原来的栈多了一个功能，可以返回当前栈内的最小值
解法：使用2个栈，栈1记录进来的数，栈2记录目前的最小值。当有新数push进来的时候，如果栈2为空或者这个数小于栈2顶上的值，就把这个数推入栈2。当pop的数正好等于最小值时，说明当前栈内的最小值变化了，要弹出这个最小值，记录的下一个最小值来到栈顶。
C++
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
s1.push(x);
if(s2.empty() || x<=s2.top())
s2.push(x);
}
void pop() {
if(s1.top()==s2.top()) s2.pop();
s1.pop();
}
int top() {
return s1.top();
}
int getMin() {
return s2.top();
}
private:
stack<int> s1,s2;
};
224. Basic Calculator
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 .
Example 1:
Input: “1 + 1”
Output: 2
Example 2:
Input: " 2-1 + 2 "
Output: 3
Example 3:
Input: “(1+(4+5+2)-3)+(6+8)”
Output: 23
JAVA(难）
class Solution {
public int calculate(String s) {
if (s.length() == 0) {
return 0;
}
s = "(" + s + ")";
int[] pos = {0};
return eval(s, pos);
}
private static int eval(String s, int[] pos) {
int val = 0, i = pos[0], sign = 1, num = 0;
while(i < s.length()) {
char c = s.charAt(i);
switch(c) {
case '+': val = val + sign * num; num = 0; sign = 1; i++; break;
case '-': val = val + sign * num; num = 0; sign = -1; i++; break;
case '(': pos[0] = i + 1; val = val + sign * eval(s, pos); i = pos[0]; break;
case ')': pos[0] = i + 1; return val + sign * num;
case ' ': i++; continue;
default : num = num * 10 + c - '0'; i++;
}
}
return val;
}
}
232. Implement Queue using Stacks
Implement the following operations of a queue using stacks.
push(x) – Push element x to the back of queue.
pop() – Removes the element from in front of queue.
peek() – Get the front element.
empty() – Return whether the queue is empty.
Example:
MyQueue queue = new MyQueue();
queue.push(1);
queue.push(2);
queue.peek(); // returns 1
queue.pop(); // returns 1
queue.empty(); // returns false
这道题让我们用栈来实现队列。栈和队列的核心不同点就是栈是先进后出，而队列是先进先出，那么我们要用栈的先进后出的特性来模拟出队列的先进先出。那么怎么做呢，其实很简单，只要我们在插入元素的时候每次都都从前面插入即可，比如如果一个队列是1,2,3,4，那么我们在栈中保存为4,3,2,1，那么返回栈顶元素1，也就是队列的首元素，则问题迎刃而解。所以此题的难度是push函数，我们需要一个辅助栈tmp，把s的元素也逆着顺序存入tmp中，此时加入新元素x，再把tmp中的元素存回来，这样就是我们要的顺序了，其他三个操作也就直接调用栈的操作即可
C++
class MyQueue {
public:
/** Initialize your data structure here. */
MyQueue() {
}
/** Push element x to the back of queue. */
void push(int x) {
stack<int> tmp;
while(!s.empty())
{
tmp.push(s.top());
s.pop();
}
s.push(x);
while(!tmp.empty())
{
s.push(tmp.top());
tmp.pop();
}
}
/** Removes the element from in front of queue and returns that element. */
int pop() {
int r=s.top();
s.pop();
return r;
}
/** Get the front element. */
int peek() {
return s.top();
}
/** Returns whether the queue is empty. */
bool empty() {
return s.empty();
}
private:
stack<int> s;
};
496. Next Greater Element I
You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the corresponding places of nums2.
The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
题目其实很简单，关键在于理解：
以nums1 = [4, 1, 2]，nums2 = [1, 3, 4, 2]为例，
nums1[0] = 4，对应为nums2[2]，在nums2中，4右边没有大于4的值，因此返回-1
nums1[1] = 1，对应为nums2[0]，在nums2中，1右边第一个大于1的值为3，因此返回3
分析：使用栈，从后往前遍历nums2[i]，每当栈不为空的时候，一直出栈直到遇到比nums2[i]大的数字停止。设立一个map<int, int> m，存储nums2中每一个元素以及它对应的下一个最大元素构成的映射。如果停止后栈为空就将m[nums2[i]]标记为-1，否则就写栈的栈顶元素～
最后将Nums1中出现的每一个元素对应的map的值放入result数组中返回～
C++
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
stack<int> s;
map<int,int> m;
for(int i=nums2.size()-1;i>=0;i--)
{
while(!s.empty() && s.top()<=nums2[i])
s.pop();
m[nums2[i]]=s.empty()?-1:s.top();
s.push(nums2[i]);
}
for(int i=0;i<nums1.size();i++)
result.push_back(m[nums1[i]]);
return result;
}
};