Commit 15c34a2

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Merge pull request youngyangyang04#2846 from Poivre-hxx/master

fix: 70 爬楼梯删去复杂度的内联公式符

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problems
- 0070.爬楼梯.md
- 0518.零钱兑换II.md

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`‎problems/0070.爬楼梯.md‎`

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Original file line number	Diff line number	Diff line change
`@@ -130,8 +130,8 @@ public:`
`130`	`130`	`};`
`131`	`131`	```
`132`	`132`
`133`		`-* 时间复杂度:$O(n)$`
`134`		`-* 空间复杂度:$O(n)$`
	`133`	`+* 时间复杂度:O(n)`
	`134`	`+* 空间复杂度:O(n)`
`135`	`135`
`136`	`136`	`当然依然也可以,优化一下空间复杂度,代码如下:`
`137`	`137`
`@@ -154,8 +154,8 @@ public:`
`154`	`154`	`};`
`155`	`155`	```
`156`	`156`
`157`		`-* 时间复杂度:$O(n)$`
`158`		`-* 空间复杂度:$O(1)$`
	`157`	`+* 时间复杂度:O(n)`
	`158`	`+* 空间复杂度:O(1)`
`159`	`159`
`160`	`160`	`后面将讲解的很多动规的题目其实都是当前状态依赖前两个,或者前三个状态,都可以做空间上的优化,但我个人认为面试中能写出版本一就够了哈,清晰明了,如果面试官要求进一步优化空间的话,我们再去优化。`
`161`	`161`
`@@ -524,3 +524,4 @@ impl Solution {`
`524`	`524`	`<a href="https://programmercarl.com/other/kstar.html" target="_blank">`
`525`	`525`	`<img src="../pics/网站星球宣传海报.jpg" width="1000"/>`
`526`	`526`	`</a>`
	`527`	`+`

`‎problems/0518.零钱兑换II.md‎`

Lines changed: 1 addition & 35 deletions

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`@@ -546,7 +546,7 @@ object Solution {`
`546`	`546`	`}`
`547`	`547`	`}`
`548`	`548`	```
`549`		`-## C`
	`549`	`+### C`
`550`	`550`
`551`	`551`	```c
`552`	`552`	`int change(int amount, int* coins, int coinsSize) {`
`@@ -593,37 +593,3 @@ public class Solution`
`593`	`593`	`<a href="https://programmercarl.com/other/kstar.html" target="_blank">`
`594`	`594`	`<img src="../pics/网站星球宣传海报.jpg" width="1000"/>`
`595`	`595`	`</a>`
`596`		`-`
`597`		`-----------`
`598`		`-`
`599`		`-`
`600`		`-`
`601`		`-回归本题,动规五步曲来分析如下:`
`602`		`-`
`603`		`-1. 确定dp数组以及下标的含义`
`604`		`-`
`605`		`-dp[j]:凑成总金额j的货币组合数为dp[j]`
`606`		`-`
`607`		`-2. 确定递推公式`
`608`		`-`
`609`		`-dp[j] 就是所有的dp[j - coins[i]](考虑coins[i]的情况)相加。`
`610`		`-`
`611`		`-所以递推公式:dp[j] += dp[j - coins[i]];`
`612`		`-`
`613`		`-这个递推公式大家应该不陌生了,我在讲解01背包题目的时候在这篇[494. 目标和](https://programmercarl.com/0494.目标和.html)中就讲解了,求装满背包有几种方法,公式都是:dp[j] += dp[j - nums[i]];`
`614`		`-`
`615`		`-3. dp数组如何初始化`
`616`		`-`
`617`		`-首先dp[0]一定要为1,dp[0] = 1是递归公式的基础。如果dp[0] = 0 的话,后面所有推导出来的值都是0了。`
`618`		`-`
`619`		`-那么 dp[0] = 1 有没有含义,其实既可以说凑成总金额0的货币组合数为1,也可以说凑成总金额0的货币组合数为0,好像都没有毛病。`
`620`		`-`
`621`		`-但题目描述中,也没明确说 amount = 0 的情况,结果应该是多少。`
`622`		`-`
`623`		`-这里我认为题目描述还是要说明一下,因为后台测试数据是默认,amount = 0 的情况,组合数为1的。`
`624`		`-`
`625`		`-下标非0的dp[j]初始化为0,这样累计加dp[j - coins[i]]的时候才不会影响真正的dp[j]`
`626`		`-`
`627`		`-dp[0]=1还说明了一种情况:如果正好选了coins[i]后,也就是j-coins[i] == 0的情况表示这个硬币刚好能选,此时dp[0]为1表示只选coins[i]存在这样的一种选法。`
`628`		`-`
`629`		`-----------------`

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`‎problems/0070.爬楼梯.md‎`

`‎problems/0518.零钱兑换II.md‎`

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