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2019-05-08 11:31:46 +08:00
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notes/Leetcode 题解 - 动态规划.md
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@ -61,7 +61,6 @@
<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
@ -93,7 +92,6 @@ public int climbStairs(int n) {
<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;
@ -206,6 +204,8 @@ public int minPathSum(int[][] grid) {
<div align="center"> <img src="pics/dc82f0f3-c1d4-4ac8-90ac-d5b32a9bd75a.jpg" width=""> </div><br>
<div align="center"> <img src="pics/dc82f0f3-c1d4-4ac8-90ac-d5b32a9bd75a.jpg"/> </div><br>
```java
public int uniquePaths(int m, int n) {
int[] dp = new int[n];
@ -979,6 +979,7 @@ public int combinationSum4(int[] nums, int target) {
<div align="center"> <img src="pics/83acbb02-872a-4178-b22a-c89c3cb60263.jpg" width="300px"> </div><br>
```java
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {