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notes/Leetcode 题解 - 动态规划.md
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@ -59,7 +59,7 @@
<!--<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"/> </div><br>
<div align="center"> <img src="pics/14fe1e71-8518-458f-a220-116003061a83.png"/> </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。
@ -91,7 +91,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/2de794ca-aa7b-48f3-a556-a0e2708cb976.jpg"/> </div><br>
```java
public int rob(int[] nums) {
@ -145,7 +145,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/da1f96b9-fd4d-44ca-8925-fb14c5733388.png"/> </div><br>
## 母牛生产
@ -157,7 +157,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/879814ee-48b5-4bcb-86f5-dcc400cb81ad.png"/> </div><br>
# 矩阵路径
@ -203,7 +203,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/dc82f0f3-c1d4-4ac8-90ac-d5b32a9bd75a.jpg"/> </div><br>
```java
public int uniquePaths(int m, int n) {
@ -400,7 +400,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/ee994da4-0fc7-443d-ac56-c08caf00a204.jpg"/> </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。
@ -567,7 +567,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/ecd89a22-c075-4716-8423-e0ba89230e9a.jpg"/> </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。
@ -607,7 +607,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/8cb2be66-3d47-41ba-b55b-319fc68940d4.png"/> </div><br>
```java
public int knapsack(int W, int N, int[] weights, int[] values) {
@ -632,7 +632,7 @@ 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"/> </div><br>
<div align="center"> <img src="pics/9ae89f16-7905-4a6f-88a2-874b4cac91f4.jpg"/> </div><br>
因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，以防将 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。
@ -976,7 +976,7 @@ 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/83acbb02-872a-4178-b22a-c89c3cb60263.jpg" width="300px"> </div><br>
```java
public int maxProfit(int[] prices) {
@ -1017,7 +1017,7 @@ 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>
<div align="center"> <img src="pics/1e2c588c-72b7-445e-aacb-d55dc8a88c29.png" width="300px"> </div><br>
```java
public int maxProfit(int[] prices, int fee) {