Commit 5982173

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feat: update solutions to lc problems

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+37

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Original file line number	Diff line number	Diff line change
`@@ -87,6 +87,23 @@ class Solution:`
`87`	`87`	`k -= 1`
`88`	`88`	```
`89`	`89`
	`90`	+```python
	`91`	`+class Solution:`
	`92`	`+ def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:`
	`93`	`+ """`
	`94`	`+ Do not return anything, modify nums1 in-place instead.`
	`95`	`+ """`
	`96`	`+ i, j, k = m - 1, n - 1, m + n - 1`
	`97`	`+ while j >= 0:`
	`98`	`+ if i < 0 or nums1[i] < nums2[j]:`
	`99`	`+ nums1[k] = nums2[j]`
	`100`	`+ j -= 1`
	`101`	`+ else:`
	`102`	`+ nums1[k] = nums1[i]`
	`103`	`+ i -= 1`
	`104`	`+ k -= 1`
	`105`	+```
	`106`	`+`
`90`	`107`	`### Java`
`91`	`108`
`92`	`109`	`<!-- 这里可写当前语言的特殊实现逻辑 -->`

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Original file line number	Diff line number	Diff line change
`@@ -76,6 +76,23 @@ class Solution:`
`76`	`76`	`k -= 1`
`77`	`77`	```
`78`	`78`
	`79`	+```python
	`80`	`+class Solution:`
	`81`	`+ def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:`
	`82`	`+ """`
	`83`	`+ Do not return anything, modify nums1 in-place instead.`
	`84`	`+ """`
	`85`	`+ i, j, k = m - 1, n - 1, m + n - 1`
	`86`	`+ while j >= 0:`
	`87`	`+ if i < 0 or nums1[i] < nums2[j]:`
	`88`	`+ nums1[k] = nums2[j]`
	`89`	`+ j -= 1`
	`90`	`+ else:`
	`91`	`+ nums1[k] = nums1[i]`
	`92`	`+ i -= 1`
	`93`	`+ k -= 1`
	`94`	+```
	`95`	`+`
`79`	`96`	`### Java`
`80`	`97`
`81`	`98`	```java

Lines changed: 2 additions & 2 deletions

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Original file line number	Diff line number	Diff line change
`@@ -76,7 +76,7 @@`
`76`	`76`
`77`	`77`	`设 $dp[i][j]$ 表示对一块高为 $i,ドル宽为 $j$ 的木块切割后能得到的最多钱数。答案就是 $dp[m][n]$。`
`78`	`78`
`79`		`-时间复杂度 $O(mn(m+n))$。`
	`79`	`+时间复杂度 $O(mn(m+n))$。`
`80`	`80`
`81`	`81`	`<!-- tabs:start -->`
`82`	`82`

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