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Commit 88da39b

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新增堆栈部分练习
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‎README.md‎

Lines changed: 4 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -84,27 +84,18 @@
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## [哈希算法](./hash.md)❌
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下面这四大基本算法,贪心,回溯,DP属于一类,这三个算法解决问题的模型都可以定义为"多阶段最优解问题"。
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<!--# 解决多阶段决策最优解模型的算法
89-
解决问题的过程中,需要经过多个决策阶段,每个决策都会对应一个状态。我们寻找一组决策序列,经过这组决策序列,能够产生最终期望的最优值。我们把这种问题模型称为<font size=5 color=red>多阶段决策最优解模型</font>. DP,回溯,贪心都可以解决这类问题.
89+
回溯算法属于万金油,基本能用DP和贪心解决的题目都能用回溯暴力解决,不过回溯算法穷举的做法使得它的复杂度非常高,是指数级别的,所以只能适用于小规模数据的问题。
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利用动态规划解决的问题,需要满足三个特征:
91+
DP比回溯更高效,但并不是所有问题都可以通过DP解决的。能用DP解决的问题,需要满足最优子结构,无后效性,重复子问题这三个特征。DP之所以如此高效,原因就在于它通过合并重复的状态,来达到剪枝的目的(比如矩阵从左上角移动到右下角,求最短路径这个题目,要么下一步向右走,要么下一步向下走,我们选择那条之前状态是最短路径的那条,这样我们就打到了剪枝的目的).
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93-
1. 最优子结构, 就是说后面的状态可以通过前面的状态推导出来
94-
2. 无后效性, 就是说一旦状态确定,就不会更改
95-
3. 重复子问题, 就是说不同的决策序列,到达某个相同的阶段时,可能会产生不同的状态。
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贪心算法实际上是DP的一种特殊情况,它能解决的问题更加有限。需要满足三个条件:
94+
贪心算法能解决的问题更加局限。需要满足三个条件,最优子结构,无后效性和贪心选择性,即通过局部最优可以产生全局的最优选择。
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1. 最优子结构
100-
2. 无后效性
101-
3. 贪心选择性, 意思就是局部最优选择,能产生全局最优选择。
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所以贪心算法能否解决算法问题的关键在于: 局部最优能不能达到全局最优?
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回溯算法是"万金油"。基本上贪心和dp能解决的问题,回溯都能解决。回溯相当于穷举搜索,列举出所有的情况,然后对比得到最优解。不过回溯的复杂度一般都是指数级的,只能用来解决小规模数据的问题。-->
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## [贪心](./greed.md)✅
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## [回溯算法](./backtracking.md)✅

‎alg-cpp.xcodeproj/project.pbxproj‎

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254255
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258+
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@@ -716,6 +722,7 @@
716722
3A04E7432341F1B7006770AF /* double11advance.h */,
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094524BA234A32D000CCBFB9 /* levenshtein_distance.h */,
718724
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09A7AC1D234E34EA00DFECF0 /* matrix_move.h */,
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@@ -826,6 +833,18 @@
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839+
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841+
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829848
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@@ -889,6 +908,7 @@
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890909
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children = (
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892912
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893913
3A5C8BA422E05ABF00354740 /* leetcode */,
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3A5C8B9622E0109200354740 /* main.cpp */,

‎dp.md‎

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@@ -11,8 +11,24 @@
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## 什么样的问题适合使用动态规划来解决?
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动态规划适合用于<font size=5 color=red>多阶段决策最优解模型</font>, 就是说动态规划一般是用来解决最优解问题,而解决问题的过程需要经历多个阶段,每个阶段决策之后会对应一组状态,然后我们寻找一组决策序列,经过这组决策序列,最终求出问题的最优解。
14+
step1: 首先看它是否属于<font size=5 color=red>多阶段决策最优解模型</font>,就是说解决问题的过程需要经历多个阶段,每个阶段决策之后会对应一组状态,然后我们寻找一组决策序列,经过这组决策序列,最终求出问题的最优解。
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16+
step2: 是否满足最优子结构,即是否能够从前面的状态推导出后面的状态。
17+
18+
step3: 是否满足无后效性,即前面状态一旦确定,就不会受之后阶段决策的影响。并且我们通过推到后面的状态的时候,只关心前面状态的值,而并不用关心它是怎么一步步推导出来的。
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20+
step4: 是否满足重复子问题,即当到达同一个阶段时是否存在多个状态.
21+
22+
23+
## 动态规划的解题思路
24+
25+
26+
关键点在于,<font size=5 color=red>如何定义状态,如何通过之前的状态推导出后面阶段的状态。</font>
27+
28+
对照着下面整个具体的例子,可以更好的去理解上面这些概念.
29+
30+
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<font size=5 color=red>[matrix_move](./dp/matrix_move.h)</font>
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1834

@@ -24,6 +40,7 @@
2440
| 双11打折优惠问题 | [double11advance](./dp/double11advance.h)|✅|
2541
| 硬币找零问题 | [coinChange2](./dp/coinChange.h)|✅|
2642
| 杨辉三角变种 | [pascals_triangle](./dp/pascals_triangle.h)|✅|
43+
| 矩阵从左上角移动到右下角 | [matrix_move](./dp/matrix_move.h)|✅|
2744
| 莱文斯坦最短编辑距离 | -- |❌|
2845
| 两个字符串的最长公共子序列 | -- |❌|
2946
| 数据序列的最长递增子序列 | -- | ❌|

‎dp/main.cpp‎

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Original file line numberDiff line numberDiff line change
@@ -14,6 +14,7 @@
1414
#include "coinChange.h"
1515
#include "levenshtein_distance.h"
1616
#include "pascals_triangle.h"
17+
#include "matrix_move.h"
1718
int main(int argc, const char * argv[]) {
1819
// insert code here...
1920
using namespace std;
@@ -22,5 +23,6 @@ int main(int argc, const char * argv[]) {
2223
test_coinChange();
2324
test_levenshtein_distance();
2425
test_pascals_triangle();
26+
test_matrix_move();
2527
return 0;
2628
}

‎dp/matrix_move.h‎

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,113 @@
1+
//
2+
// matrix_move.h
3+
// dp
4+
//
5+
// Created by junl on 2019年10月9日.
6+
// Copyright © 2019 junl. All rights reserved.
7+
//
8+
9+
#ifndef matrix_move_hpp
10+
#define matrix_move_hpp
11+
12+
#include <stdio.h>
13+
#include <vector>
14+
/*
15+
和前面的杨辉三角变种有点类似:
16+
问题是有一个n*n的矩阵,矩阵存储的都是正整数,旗子最开始位于左上角,终止位置在右下角。中间经过的数字是它们的路径和,求最短的路径和.
17+
18+
1,3,5,9
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2,1,3,4
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5,2,6,7
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6,8,4,3
22+
*/
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using namespace std;
24+
class matrix_move {
25+
public:
26+
int solve(vector<vector<int>> &matrix){
27+
if (matrix.empty() || matrix.size() != matrix[0].size()) {
28+
return 0;
29+
}
30+
solve(matrix, 0, 0, matrix[0][0]);
31+
return result;
32+
}
33+
private:
34+
int result = INT_MAX;
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36+
void solve(vector<vector<int>> &matrix, int row,int col ,int pathLen){
37+
size_t N = matrix.size()-1;
38+
if (row == N && col == N) {
39+
result = min(result, pathLen);
40+
return;
41+
}
42+
if (row < N) {
43+
solve(matrix, row+1, col, pathLen+matrix[row+1][col]);
44+
}
45+
if (col < N) {
46+
solve(matrix, row, col+1, pathLen+matrix[row][col+1]);
47+
}
48+
}
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};
50+
51+
52+
class matrix_move_dp {
53+
public:
54+
/*
55+
思路:
56+
假设矩阵大小为N*N,那么从左上角移动到右下角一共要移动2*(N-1)次.
57+
每一次要么向下移动,要么向右移动,每次移动之后,状态(路径和)都会发生变化。该问题符合动态规划的"一个模型三个条件"
58+
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怎么定义状态呢?
60+
-- int min_path[i][j] 里面保存的是从左上角到matrix[i][j]的最小路径和
61+
状态转移方程是什么?
62+
-- 走法只有两种,那么到达matrix[i][j]的位置,只有可能是min_path[i-1][j]或者min_path[i][j-1]中的一条.
63+
-- min_path[i][j] = min(min_path[i-1][j], min_path[i][j-1]) + matrix[i][j];
64+
*/
65+
int solve(vector<vector<int>> &matrix){
66+
if (matrix.empty()) {
67+
return 0;
68+
}
69+
size_t N = matrix.size();
70+
int min_path[N][N];
71+
memset(min_path, 0, sizeof(min_path));
72+
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//设置初始状态
74+
int sum=0;
75+
int i;
76+
for (i=0; i<N; i++) {//第一排的初始状态
77+
sum+=matrix[0][i];
78+
min_path[0][i]=sum;
79+
}
80+
81+
sum=0;
82+
for (i=0; i<N; i++) { //第一列的初始状态
83+
sum+=matrix[i][0];
84+
min_path[i][0]=sum;
85+
}
86+
87+
//通过初始状态动态向前推导,按照一行行的向下推导
88+
for (i=1; i<N; i++) {
89+
for (int j=1; j<N; j++) {
90+
min_path[i][j] = min(min_path[i-1][j], min_path[i][j-1]) + matrix[i][j];
91+
}
92+
}
93+
return min_path[N-1][N-1];
94+
};
95+
};
96+
97+
void test_matrix_move(){
98+
vector<vector<int>> matrix;
99+
matrix.push_back({1,3,5,9});
100+
matrix.push_back({2,1,3,4});
101+
matrix.push_back({5,2,6,7});
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matrix.push_back({6,8,4,3});
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class matrix_move so;
104+
cout << "----------矩阵移动-----------" << endl;
105+
cout << so.solve(matrix) << endl;
106+
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class matrix_move_dp so_dp;
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cout << so_dp.solve(matrix) << endl;
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}
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#endif /* matrix_move_hpp */
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,70 @@
1+
//
2+
// TwoStacksQueue.h
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// stack+queue
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//
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// Created by junl on 2019年10月14日.
6+
// Copyright © 2019 junl. All rights reserved.
7+
//
8+
9+
#ifndef TwoStacksQueue_hpp
10+
#define TwoStacksQueue_hpp
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#include <stdio.h>
13+
#include <stack>
14+
/*
15+
思路:
16+
两个栈正好可以实现倒序.关键点把pushStack的数据倒序放到popStack中:
17+
1.如果popStack包含元素了的话,那么不能将数据从pushStack移动到popStack
18+
2.必须一次性的把pushStackd里面的元素移动到popStack中
19+
*/
20+
class TwoStacksQueue {
21+
public:
22+
typedef int Value_t;
23+
void add(Value_t x)//添加元素
24+
{
25+
pushStack.push(x);
26+
}
27+
void poll()//删除队列首元素
28+
{
29+
move();
30+
if (popStack.empty()) {
31+
throw stackEmpty();
32+
}
33+
popStack.pop();
34+
}
35+
Value_t peek()//获取首元素
36+
{
37+
if (popStack.empty() && pushStack.empty()) {
38+
throw stackEmpty();
39+
}
40+
move();
41+
return popStack.top();
42+
}
43+
private:
44+
std::stack<Value_t> pushStack;
45+
std::stack<Value_t> popStack;
46+
void move(){
47+
if (popStack.empty()) {
48+
//如果popStack为空,则把pushStack的所有元素倒序加入.
49+
while (!pushStack.empty()) {
50+
popStack.push(pushStack.top());
51+
pushStack.pop();
52+
}
53+
}
54+
}
55+
};
56+
57+
58+
void test_TwoStacksQueue(){
59+
std::cout << "-----------test_TwoStacksQueue-----------" << std::endl;
60+
TwoStacksQueue queue;
61+
queue.add(1);
62+
queue.add(2);
63+
queue.add(3);
64+
std::cout << "peek: "<< queue.peek() << std::endl;//1
65+
queue.poll();
66+
std::cout << "peek: "<< queue.peek() << std::endl;//2
67+
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}
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#endif /* TwoStacksQueue_hpp */

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