剑指offerWeek3

周五：顺时针打印矩阵

题目链接：顺时针打印矩阵

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。数据范围
矩阵中元素数量 [0,400]样例
输入：
[[1, 2, 3, 4],[5, 6, 7, 8],[9,10,11,12]
]输出：[1,2,3,4,8,12,11,10,9,5,6,7]

AC代码

class Solution {
public:vector<int> printMatrix(vector<vector<int> > matrix) {vector<int> res;if (matrix.empty()) return res;int n = matrix.size(), m = matrix[0].size();int x = 0, y = 0, d = 1;int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};vector<vector<bool>> st(n, vector<bool>(m + 1, false));for (int i = 0; i < n * m; i ++ ){res.push_back(matrix[x][y]);st[x][y] = true;int a = dx[d] + x, b = dy[d] + y;if (a < 0 || b < 0 || a >= n || b >= m || st[a][b]){d = (d + 1) % 4;a = dx[d] + x, b = dy[d] + y;}x = a, y = b;}return res;}
};

思路：

整体思路

简单深搜罢了可以背记一下遍历模板int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
for (int i = 0; i < 4; i ++ )
{int a = x + dx[i], b = dy[i] + y;if (a < 0 && b < 0 && a < n && b < m && (其他边界条件)){//}
}

周五：包含min函数的栈

题目链接：包含min函数的栈

设计一个支持push，pop，top等操作并且可以在O(1)时间内检索出最小元素的堆栈。push(x)–将元素x插入栈中
pop()–移除栈顶元素
top()–得到栈顶元素
getMin()–得到栈中最小元素
数据范围
操作命令总数 [0,100]样例
MinStack minStack = new MinStack();
minStack.push(-1);
minStack.push(3);
minStack.push(-4);
minStack.getMin();   --> Returns -4.
minStack.pop();
minStack.top();      --> Returns 3.
minStack.getMin();   --> Returns -1.

AC代码

class MinStack {
public:/** initialize your data structure here. */stack<int> valStk;stack<int> minStk;MinStack() {}void push(int x) {valStk.push(x);if (minStk.empty() || minStk.top() >= x) minStk.push(x);}void pop() {if (minStk.top() == valStk.top()) minStk.pop();valStk.pop();}int top() {return valStk.top();}int getMin() {return minStk.top();}
};/*** Your MinStack object will be instantiated and called as such:* MinStack obj = new MinStack();* obj.push(x);* obj.pop();* int param_3 = obj.top();* int param_4 = obj.getMin();*/

思路：

整体思路

单独维护一个栈即可
有点类似单调栈