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docs/notes/Leetcode 题解 - 动态规划.md
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@ -904,7 +904,7 @@ Explanation: there are four ways to make up the amount:
```java
public int change(int amount, int[] coins) {
if (amount == 0 || coins == null || coins.length == 0) {
if (coins == null) {
return 0;
}
int[] dp = new int[amount + 1];

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notes/Leetcode 题解 - 动态规划.md
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@ -57,6 +57,7 @@
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/14fe1e71-8518-458f-a220-116003061a83.png" width="200px"> </div><br>
考虑到 dp[i] 只与 dp[i - 1] 和 dp[i - 2] 有关，因此可以只用两个变量来存储 dp[i - 1] 和 dp[i - 2]，使得原来的 O(N) 空间复杂度优化为 O(1) 复杂度。
```java
@ -87,6 +88,7 @@ public int climbStairs(int n) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2]+nums[i],dp[i-1])" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/2de794ca-aa7b-48f3-a556-a0e2708cb976.jpg" width="350px"> </div><br>
```java
public int rob(int[] nums) {
int pre2 = 0, pre1 = 0;
@ -140,6 +142,7 @@ private int rob(int[] nums, int first, int last) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=(i-1)*dp[i-2]+(i-1)*dp[i-1]" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/da1f96b9-fd4d-44ca-8925-fb14c5733388.png" width="350px"> </div><br>
## 5. 母牛生产
[程序员代码面试指南-P181](#)
@ -151,6 +154,7 @@ private int rob(int[] nums, int first, int last) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/879814ee-48b5-4bcb-86f5-dcc400cb81ad.png" width="250px"> </div><br>
# 矩阵路径
## 1. 矩阵的最小路径和
@ -196,6 +200,7 @@ public int minPathSum(int[][] grid) {
题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。
<div align="center"> <img src="pics/dc82f0f3-c1d4-4ac8-90ac-d5b32a9bd75a.jpg" width=""> </div><br>
```java
public int uniquePaths(int m, int n) {
int[] dp = new int[n];
@ -416,6 +421,7 @@ public int numDecodings(String s) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[n]=max\{1,dp[i]+1|S_i<S_n\&\&i<n\}" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/ee994da4-0fc7-443d-ac56-c08caf00a204.jpg" width="350px"> </div><br>
对于一个长度为 N 的序列，最长递增子序列并不一定会以 S<sub>N</sub> 为结尾，因此 dp[N] 不是序列的最长递增子序列的长度，需要遍历 dp 数组找出最大值才是所要的结果，max{ dp[i] | 1 <= i <= N} 即为所求。
## 1. 最长递增子序列
@ -582,6 +588,7 @@ public int wiggleMaxLength(int[] nums) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=\left\{\begin{array}{rcl}dp[i-1][j-1]&&{S1_i==S2_j}\\max(dp[i-1][j],dp[i][j-1])&&{S1_i<>S2_j}\end{array}\right." class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/ecd89a22-c075-4716-8423-e0ba89230e9a.jpg" width="450px"> </div><br>
对于长度为 N 的序列 S<sub>1</sub> 和长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
与最长递增子序列相比，最长公共子序列有以下不同点：
@ -621,6 +628,7 @@ public int lengthOfLCS(int[] nums1, int[] nums2) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=max(dp[i-1][j],dp[i-1][j-w]+v)" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/8cb2be66-3d47-41ba-b55b-319fc68940d4.png" width="400px"> </div><br>
```java
// W 为背包总体积
// N 为物品数量
@ -649,6 +657,7 @@ public int knapsack(int W, int N, int[] weights, int[] values) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[j]=max(dp[j],dp[j-w]+v)" class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/9ae89f16-7905-4a6f-88a2-874b4cac91f4.jpg" width="300px"> </div><br>
因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，防止将 dp[i-1][j-w] 覆盖。也就是说要先计算 dp[i][j] 再计算 dp[i][j-w]，在程序实现时需要按倒序来循环求解。
```java
@ -860,17 +869,20 @@ return -1.
```java
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i < dp.length; i++) {
for (int j = 0; j < coins.length; j++) {
if (coins[j] <= i) {
dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
}
int[] dp = new int[amount + 1];
for (int coin : coins) {
for (int i = coin; i <= amount; i++) { //将逆序遍历改为正序遍历
if (i == coin) {
dp[i] = 1;
} else if (dp[i] == 0 && dp[i - coin] != 0) {
dp[i] = dp[i - coin] + 1;
} else if (dp[i - coin] != 0) {
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
return dp[amount] == 0 ? -1 : dp[amount];
}
```
@ -892,7 +904,7 @@ Explanation: there are four ways to make up the amount:
```java
public int change(int amount, int[] coins) {
if (amount == 0 || coins == null || coins.length == 0) {
if (coins == null) {
return 0;
}
int[] dp = new int[amount + 1];
@ -992,8 +1004,7 @@ public int combinationSum4(int[] nums, int target) {
题目描述：交易之后需要有一天的冷却时间。
<div align="center"> <img src="pics/83acbb02-872a-4178-b22a-c89c3cb60263.jpg" width="300px"> </div><br>
<div align="center"> <img src="pics/ffd96b99-8009-487c-8e98-11c9d44ef14f.png" width="300px"> </div><br>
```java
public int maxProfit(int[] prices) {
@ -1035,6 +1046,7 @@ The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
题目描述：每交易一次，都要支付一定的费用。
<div align="center"> <img src="pics/1e2c588c-72b7-445e-aacb-d55dc8a88c29.png" width="300px"> </div><br>
```java
public int maxProfit(int[] prices, int fee) {
int N = prices.length;