| Difficulty | topic | Knowledge Point |
|---|---|---|
| 07. Fibonacci number | Recursive recursive - written two variables - | |
| 08. jump stairs | Ditto | |
| 09. metamorphosis jump stairs | dp | |
| 10. The rectangular footprint | Ditto | |
| 05. queue with two stacks | simulation | |
| ☆ | 20. The stack includes a function min | Stack |
| 21. The press-pop stack sequence | Stack simulated sequence | |
| 65. The path matrix | Backtracking | |
| 66. The range of motion of the robot | dfs communication request block size |
07--10 Fibonacci number - recursive recursive - two variables wording
07. Fibonacci number
T7: We all know that Fibonacci number, and now asked to enter an integer n, you output the n-th Fibonacci number Fibonacci sequence of item (from 0, the first 0 is 0). n <= 39.
class Solution {
public:
int Fibonacci(int n) {
if(n==0) return 0;
if(n==1) return 1;
if(n==2) return 1;
return Fibonacci(n-1)+Fibonacci(n-2);
}
};08. jump stairs
T8: a frog can jump on a Class 1 level, you can also hop on level 2. The frog jumped seeking a total of n grade level how many jumps (the order of different calculation different results).
// 两变量
class Solution {
public:
int jumpFloor(int number) {
if(number==0)return 1;
if(number==1)return 1;
if(number==2)return 2;
int f=1,g=2;
number-=2;
while(number--){
g=f+g;
f=g-f;
}
return g;
}
};09. metamorphosis jump stairs
T9: a frog can jump on a Class 1 level, you can also hop on level 2 ...... n It can also jump on stage. The frog jumped seeking a total of n grade level how many jumps.
// dp
class Solution {
public:
int jumpFloorII(const int number) {
int** dp=new int*[number+10];
for(int i=0;i<number+10;i++){
dp[i]=new int[number+10];
}
memset(dp,0,sizeof dp);
for(int i=1;i<=number;i++){
dp[1][i]=1;
}
// dp[i][j] 用i步跳上台阶j
for(int i=2;i<=number;i++){
for(int j=i;j<=number;j++){
for(int k=i-1;k<j;k++){
dp[i][j]+=dp[i-1][k];
}
}
}
int ans=0;
for(int i=1;i<=number;i++){
ans+=dp[i][number];
}
return ans;// 返回的变量打错，不可原谅，，
}
};10. The rectangular footprint
T10: We can use a small rectangle 2 * 1 sideways or vertically to cover a larger rectangle. Will the small rectangle of n 2 * 1 coverage without overlap a large rectangle 2 * n, a total of how many ways?
Similar to the previous few questions.
class Solution {
public:
int rectCover(int number) {
if(number==0) return 0;
if(number==1) return 1;
if(number==2) return 2;
int f=1,g=2;
number-=2;
while(number--){
g=f+g;
f=g-f;
}
return g;
}
};05. queue with two stacks
Two stacks to achieve a queue, the completion queue Push and Pop operations. Queue elements int.
Received into queue element stack 1, stack 2 out of the storage elements of the queue, when the stack 2 is empty, the stack element 1 in the stack down to 2.
class Solution
{
public:
void push(int node) {
stack1.push(node);
}
int pop() {
if(stack2.size()==0){
while(!stack1.empty()){
int x=stack1.top();
stack1.pop();
stack2.push(x);
}
}
int x=stack2.top();
stack2.pop();
return x;
}
private:
stack<int> stack1;
stack<int> stack2;
};20. The stack includes a function min
Stack
Title Description
Stack data structure definition, implement this type can be a min function smallest elements contained in the stack (should the time complexity O (1)).
Each one pair of press-fitting elements (element current and the minimum current).
Java Code
import java.util.Stack;
public class Solution {
private Stack
st=new Stack<>();
public void push(int node) {
int min=node;
if(!st.empty()) min=Math.min(min,st.peek());
st.push(node);
st.push(min);
}
public void pop() {
st.pop();
st.pop();
}
public int top() {
int x=st.peek();
st.pop();
int y=st.peek();
st.push(x);
return y;
}
public int min() {
return st.peek();
}
}
More space-saving approach is as follows:
Application of an auxiliary stack when the pressure, if the pressure of the stack A is greater than B is pressed into the stack, the stack is not pressed B ,,,, less, AB pressed while the stack, the stack, if , AB ranging from the top element, A a, B no.
21. The stack is pressed into the pop-up sequence
simulation
Title Description
Two input sequence of integers, the first sequence representing a pressed stack order, determines whether it is possible for the second sequence the order of the pop-up stack. All figures are not pushed onto the stack is assumed equal. Such as a sequence of a sequence of 1,2,3,4,5 is pressed into the stack, the push sequence is 4,5,3,2,1 sequence corresponding to a sequence of pop, but 4,3,5,1,2 it is impossible to push pop-up sequence of the sequence. (Note: the length of the two sequences are equal)
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
public boolean IsPopOrder(int [] pushA,int [] popA) {
if(pushA.length == 0 || popA.length == 0)
return false;
Stack<Integer> s = new Stack<Integer>();
//用于标识弹出序列的位置
int popIndex = 0;
for(int i = 0; i< pushA.length;i++){
s.push(pushA[i]);
//如果栈不为空，且栈顶元素等于弹出序列
while(!s.empty() &&s.peek() == popA[popIndex]){
//出栈
s.pop();
//弹出序列向后一位
popIndex++;
}
}
return s.empty();
}
}65. The matrix of path
Backtracking
Title Description
Please design a function that determines whether there is a route that contains all the characters of a character string in a matrix. A lattice path may start from any of the matrix, each step in the matrix can be left, right, up, down one grid. If a route that passes through a matrix of a grid, the path can not enter the lattice. E.g. abcesfcsdee matrix containing the path "bcced" in a string, but does not contain the matrix "abcb" path, as a character string in the first row of the matrix b occupy the first lattice after the second path can not be again access to the grid.
dfs back.
Java Code
public class Solution {
private int rows, cols;
char[][] G;
char[] str;
boolean[][] vis;
int[] dx = new int[]{1, 0, -1, 0};
int[] dy = new int[]{0, 1, 0, -1};
public boolean hasPath(char[] matrix, int rows, int cols, char[] str) {
if (matrix == null || matrix.length < str.length)
return false;
this.str = str;
this.rows = rows;
this.cols = cols;
G = new char[rows][cols];
vis = new boolean[rows][cols];
for (int i = 0; i < matrix.length; i++) {
G[i / cols][i % cols] = matrix[i];
}
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (G[i][j] == str[0]) {
if (str.length == 1) return true;
vis[i][j] = true;
for (int k = 0; k < 4; k++) {
if (check(i + dx[k], j + dy[k])
&& dfs(i + dx[k], j + dy[k], 1))
return true;
}
vis[i][j] = false;
}
}
}
return false;
}
private boolean dfs(int x, int y, int idx) {
if (G[x][y] != str[idx]) return false;
if (idx == str.length - 1) return true;
vis[x][y] = true;
for (int k = 0; k < 4; k++) {
if (check(x + dx[k], y + dy[k])
&& dfs(x + dx[k], y + dy[k], idx + 1))
return true;
}
vis[x][y] = false;
return false;
}
private boolean check(int x, int y) {
return x >= 0 && x < rows && y >= 0 && y < cols && !vis[x][y];
}
}
66. The range of motion of the robot
dfs communication request block size
Title Description
ground there is a grid of m rows and n columns. A robot moves from the grid coordinates 0,0, every time only left, right, upper, lower four directions a cell, but can not enter the row and column coordinates of the grid is greater than the sum of the number of bits k. For example, when k is 18, the robot can enter the box (35, 37), because 3 + 3 + 5 + 7 = 18. However, it can not enter the box (35, 38), because 3 + 3 + 5 + 8 = 19. Will the robot be able to reach the number of lattice?
dfs communication request block size. Note that use array vis avoid double-counting.
Java Code
public class Solution {
int cnt=0;
boolean [][]vis;
public int movingCount(int threshold, int rows, int cols)
{
vis=new boolean[rows][cols];
dfs(0,0,rows,cols,threshold);
return cnt;
}
private void dfs(int x,int y,int rows,int cols,int thrsh){
if(!check(x,y,rows,cols,thrsh)) return;
cnt++;
vis[x][y]=true;
dfs(x,y+1,rows,cols,thrsh);
dfs(x+1,y,rows,cols,thrsh);
dfs(x,y-1,rows,cols,thrsh);
dfs(x-1,y,rows,cols,thrsh);
}
private boolean check(int x,int y,int rows,int cols,int thrsh){
if(x<0||x>=rows||y<0||y>=cols||vis[x][y]) return false;
int sum=0;
while(x>0){ sum+=x%10;x/=10; }
while(y>0){ sum+=y%10;y/=10; }
return sum<=thrsh;
}
}