Merge branch 'master' of github.com:youngyangyang04/leetcode-master

This commit is contained in:
youngyangyang04
2022-03-01 17:17:08 +08:00
3 changed files with 96 additions and 0 deletions

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@ -106,6 +106,37 @@ public:
旧文链接:[数组:就移除个元素很难么?](https://programmercarl.com/0027.移除元素.html) 旧文链接:[数组:就移除个元素很难么?](https://programmercarl.com/0027.移除元素.html)
```CPP
/**
* 相向双指针方法,基于元素顺序可以改变的题目描述改变了元素相对位置,确保了移动最少元素
* 时间复杂度:$O(n)$
* 空间复杂度:$O(1)$
*/
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
int leftIndex = 0;
int rightIndex = nums.size() - 1;
while (leftIndex <= rightIndex) {
// 找左边等于val的元素
while (leftIndex <= rightIndex && nums[leftIndex] != val){
++leftIndex;
}
// 找右边不等于val的元素
while (leftIndex <= rightIndex && nums[rightIndex] == val) {
-- rightIndex;
}
// 将右边不等于val的元素覆盖左边等于val的元素
if (leftIndex < rightIndex) {
nums[leftIndex++] = nums[rightIndex--];
}
}
return leftIndex; // leftIndex一定指向了最终数组末尾的下一个元素
}
};
```
## 相关题目推荐 ## 相关题目推荐
* 26.删除排序数组中的重复项 * 26.删除排序数组中的重复项

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@ -433,6 +433,51 @@ var findBottomLeftValue = function(root) {
}; };
``` ```
## TypeScript
> 递归法:
```typescript
function findBottomLeftValue(root: TreeNode | null): number {
function recur(root: TreeNode, depth: number): void {
if (root.left === null && root.right === null) {
if (depth > maxDepth) {
maxDepth = depth;
resVal = root.val;
}
return;
}
if (root.left !== null) recur(root.left, depth + 1);
if (root.right !== null) recur(root.right, depth + 1);
}
let maxDepth: number = 0;
let resVal: number = 0;
if (root === null) return resVal;
recur(root, 1);
return resVal;
};
```
> 迭代法:
```typescript
function findBottomLeftValue(root: TreeNode | null): number {
let helperQueue: TreeNode[] = [];
if (root !== null) helperQueue.push(root);
let resVal: number = 0;
let tempNode: TreeNode;
while (helperQueue.length > 0) {
resVal = helperQueue[0].val;
for (let i = 0, length = helperQueue.length; i < length; i++) {
tempNode = helperQueue.shift()!;
if (tempNode.left !== null) helperQueue.push(tempNode.left);
if (tempNode.right !== null) helperQueue.push(tempNode.right);
}
}
return resVal;
};
```
## Swift ## Swift
递归版本: 递归版本:

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@ -310,6 +310,26 @@ class Solution:
return root return root
``` ```
**递归法** - 无返回值 - another easier way
```python
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
newNode = TreeNode(val)
if not root: return newNode
if not root.left and val < root.val:
root.left = newNode
if not root.right and val > root.val:
root.right = newNode
if val < root.val:
self.insertIntoBST(root.left, val)
if val > root.val:
self.insertIntoBST(root.right, val)
return root
```
**迭代法** **迭代法**
与无返回值的递归函数的思路大体一致 与无返回值的递归函数的思路大体一致
```python ```python