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feat: add go solution to lc problem: No.0444 (doocs#868)

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`@@ -58,6 +58,8 @@`
`58`	`58`	`- 但可以选择性丢弃前面的相邻元素,丢弃与否取决于当前元素和前面相邻元素的大小;`
`59`	`59`	`- 根据前置知识可知当当前元素小于前面相邻元素时可以移除前面相邻的元素。`
`60`	`60`
	`61`	`+时间复杂度 $O(n),ドル空间复杂度 $O(n)$。`
	`62`	`+`
`61`	`63`	`<!-- tabs:start -->`
`62`	`64`
`63`	`65`	`### Python3`

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`@@ -48,13 +48,15 @@`
`48`	`48`	`动态规划转移方程如下:`
`49`	`49`
`50`	`50`	`$$`
`51`		`-dp[i][k] = dp[j][k-1] \|\| dp[j][k] \|\| dp[j][k+1]`
	`51`	`+dp[i][k] = dp[j][k-1] \ \|\| \ dp[j][k] \ \|\| \ dp[j][k+1]`
`52`	`52`	`$$`
`53`	`53`
`54`	`54`	其中 `dp[i][k]` 表示最后一次跳跃为 `k` 个单位时,能否到达 `i`,定义 base case 为 `dp[0][0] = True`(起点在下标 0)。
`55`	`55`
`56`	`56`	对于从 `j` 跳到 `i` 的青蛙,因为跳跃的距离确定为 `k` 个单位,所以根据题意最后一次跳跃到达 `j` 的跳跃距离只能选择为 `k - 1`、`k` 或 `k + 1` 个单位,故只要 `dp[j][k - 1], dp[j][k], dp[j][k + 1]` 中有任一为 `True`,即可从 `j` 跳跃到 `i`。
`57`	`57`
	`58`	`+时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。`
	`59`	`+`
`58`	`60`	`方法二:回溯+剪枝`
`59`	`61`
`60`	`62`	这是最直观的解题思路。显然青蛙在第 `1` 个石子的起始跳跃距离为 `1`,对于第 `2` 个石子,根据题意很容易得到青蛙的跳跃距离只能是 `0、1 或 2`。依次类推,可以得到青蛙在第 `i` 个石子可能的跳跃距离集合,借助这个思路验证当青蛙在 `i` 处跳跃距离为集合之一时是否可以刚好过河,如不能过河继续验证其他取值即可。

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`@@ -189,6 +189,33 @@ impl Solution {`
`189`	`189`	`}`
`190`	`190`	```
`191`	`191`
	`192`	`+### Go`
	`193`	`+`
	`194`	`+<!-- 这里可写当前语言的特殊实现逻辑 -->`
	`195`	`+`
	`196`	+```go
	`197`	`+/**`
	`198`	`+ * Definition for a binary tree node.`
	`199`	`+ * type TreeNode struct {`
	`200`	`+ * Val int`
	`201`	`+ * Left *TreeNode`
	`202`	`+ * Right *TreeNode`
	`203`	`+ * }`
	`204`	`+ */`
	`205`	`+func sumOfLeftLeaves(root *TreeNode) int {`
	`206`	`+ if root == nil {`
	`207`	`+ return 0`
	`208`	`+ }`
	`209`	`+ res := 0`
	`210`	`+ if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {`
	`211`	`+ res += root.Left.Val`
	`212`	`+ }`
	`213`	`+ res += sumOfLeftLeaves(root.Left)`
	`214`	`+ res += sumOfLeftLeaves(root.Right)`
	`215`	`+ return res`
	`216`	`+}`
	`217`	+```
	`218`	`+`
`192`	`219`	`### ...`
`193`	`220`
`194`	`221`	```

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`@@ -177,6 +177,31 @@ impl Solution {`
`177`	`177`	`}`
`178`	`178`	```
`179`	`179`
	`180`	`+### Go`
	`181`	`+`
	`182`	+```go
	`183`	`+/**`
	`184`	`+ * Definition for a binary tree node.`
	`185`	`+ * type TreeNode struct {`
	`186`	`+ * Val int`
	`187`	`+ * Left *TreeNode`
	`188`	`+ * Right *TreeNode`
	`189`	`+ * }`
	`190`	`+ */`
	`191`	`+func sumOfLeftLeaves(root *TreeNode) int {`
	`192`	`+ if root == nil {`
	`193`	`+ return 0`
	`194`	`+ }`
	`195`	`+ res := 0`
	`196`	`+ if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {`
	`197`	`+ res += root.Left.Val`
	`198`	`+ }`
	`199`	`+ res += sumOfLeftLeaves(root.Left)`
	`200`	`+ res += sumOfLeftLeaves(root.Right)`
	`201`	`+ return res`
	`202`	`+}`
	`203`	+```
	`204`	`+`
`180`	`205`	`### ...`
`181`	`206`
`182`	`207`	```

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`@@ -0,0 +1,20 @@`
	`1`	`+/**`
	`2`	`+ * Definition for a binary tree node.`
	`3`	`+ * type TreeNode struct {`
	`4`	`+ * Val int`
	`5`	`+ * Left *TreeNode`
	`6`	`+ * Right *TreeNode`
	`7`	`+ * }`
	`8`	`+ */`
	`9`	`+func sumOfLeftLeaves(root *TreeNode) int {`
	`10`	`+ if root == nil {`
	`11`	`+ return 0`
	`12`	`+ }`
	`13`	`+ res := 0`
	`14`	`+ if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {`
	`15`	`+ res += root.Left.Val`
	`16`	`+ }`
	`17`	`+ res += sumOfLeftLeaves(root.Left)`
	`18`	`+ res += sumOfLeftLeaves(root.Right)`
	`19`	`+ return res`
	`20`	`+}`

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