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Commit a20ac9d

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Merge branch 'youngyangyang04:master' into master
2 parents 4defb3d + bda7b41 commit a20ac9d

6 files changed

+136
-27
lines changed

‎problems/0300.最长上升子序列.md

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@@ -168,6 +168,39 @@ func lengthOfLIS(nums []int ) int {
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}
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```
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```go
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// 动态规划求解
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func lengthOfLIS(nums []int) int {
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// dp数组的定义 dp[i]表示取第i个元素的时候,表示子序列的长度,其中包括 nums[i] 这个元素
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dp := make([]int, len(nums))
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// 初始化,所有的元素都应该初始化为1
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for i := range dp {
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dp[i] = 1
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}
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ans := dp[0]
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for i := 1; i < len(nums); i++ {
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for j := 0; j < i; j++ {
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if nums[i] > nums[j] {
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dp[i] = max(dp[i], dp[j] + 1)
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}
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}
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if dp[i] > ans {
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ans = dp[i]
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}
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}
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return ans
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}
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```
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Javascript
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```javascript
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const lengthOfLIS = (nums) => {

‎problems/0674.最长连续递增序列.md

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@@ -236,6 +236,45 @@ class Solution:
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```
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Go:
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> 动态规划:
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```go
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func findLengthOfLCIS(nums []int) int {
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if len(nums) == 0 {return 0}
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res, count := 1, 1
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for i := 0; i < len(nums)-1; i++ {
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if nums[i+1] > nums[i] {
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count++
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}else {
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count = 1
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}
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if count > res {
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res = count
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}
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}
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return res
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}
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```
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> 贪心算法:
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```go
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func findLengthOfLCIS(nums []int) int {
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if len(nums) == 0 {return 0}
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dp := make([]int, len(nums))
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for i := 0; i < len(dp); i++ {
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dp[i] = 1
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}
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res := 1
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for i := 0; i < len(nums)-1; i++ {
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if nums[i+1] > nums[i] {
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dp[i+1] = dp[i] + 1
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}
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if dp[i+1] > res {
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res = dp[i+1]
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}
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}
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return res
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}
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```
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Javascript:
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‎problems/0968.监控二叉树.md

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@@ -476,7 +476,35 @@ var minCameraCover = function(root) {
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};
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```
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### TypeScript
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```typescript
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function minCameraCover(root: TreeNode | null): number {
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/** 0-无覆盖, 1-有摄像头, 2-有覆盖 */
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type statusCode = 0 | 1 | 2;
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let resCount: number = 0;
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if (recur(root) === 0) resCount++;
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return resCount;
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function recur(node: TreeNode | null): statusCode {
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if (node === null) return 2;
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const left: statusCode = recur(node.left),
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right: statusCode = recur(node.right);
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let resStatus: statusCode = 0;
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if (left === 0 || right === 0) {
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resStatus = 1;
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resCount++;
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} else if (left === 1 || right === 1) {
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resStatus = 2;
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} else {
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resStatus = 0;
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}
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return resStatus;
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}
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};
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```
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### C
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```c
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/*
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**函数后序遍历二叉树。判断一个结点状态时,根据其左右孩子结点的状态进行判断

‎problems/二叉树的迭代遍历.md

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@@ -11,9 +11,9 @@
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看完本篇大家可以使用迭代法,再重新解决如下三道leetcode上的题目:
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* 144.二叉树的前序遍历
15-
* 94.二叉树的中序遍历
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* 145.二叉树的后序遍历
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* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/)
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* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/)
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* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/)
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为什么可以用迭代法(非递归的方式)来实现二叉树的前后中序遍历呢?
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‎problems/二叉树的递归遍历.md

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@@ -99,9 +99,9 @@ void traversal(TreeNode* cur, vector<int>& vec) {
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此时大家可以做一做leetcode上三道题目,分别是:
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* 144.二叉树的前序遍历
103-
* 145.二叉树的后序遍历
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* 94.二叉树的中序遍历
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* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/)
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* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/)
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* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/)
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可能有同学感觉前后中序遍历的递归太简单了,要打迭代法(非递归),别急,我们明天打迭代法,打个通透!
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‎problems/背包理论基础01背包-1.md

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### javascript
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```js
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function testweightbagproblem (wight, value, size) {
384-
const len = wight.length,
385-
dp = array.from({length: len + 1}).map(
386-
() => array(size + 1).fill(0)
387-
);
388-
389-
for(let i = 1; i <= len; i++) {
390-
for(let j = 0; j <= size; j++) {
391-
if(wight[i - 1] <= j) {
392-
dp[i][j] = math.max(
393-
dp[i - 1][j],
394-
value[i - 1] + dp[i - 1][j - wight[i - 1]]
395-
)
396-
} else {
397-
dp[i][j] = dp[i - 1][j];
398-
}
399-
}
400-
}
401-
402-
// console.table(dp);
383+
/**
384+
*
385+
* @param {Number []} weight
386+
* @param {Number []} value
387+
* @param {Number} size
388+
* @returns
389+
*/
390+
391+
function testWeightBagProblem(weight, value, size) {
392+
const len = weight.length,
393+
dp = Array.from({length: len}).map(
394+
() => Array(size + 1)) //JavaScript 数组是引用类型
395+
for(let i = 0; i < len; i++) { //初始化最左一列,即背包容量为0时的情况
396+
dp[i][0] = 0;
397+
}
398+
for(let j = 1; j < size+1; j++) { //初始化第0行, 只有一件物品的情况
399+
if(weight[0] <= j) {
400+
dp[0][j] = value[0];
401+
} else {
402+
dp[0][j] = 0;
403+
}
404+
}
405+
406+
for(let i = 1; i < len; i++) { //dp[i][j]由其左上方元素推导得出
407+
for(let j = 1; j < size+1; j++) {
408+
if(j < weight[i]) dp[i][j] = dp[i - 1][j];
409+
else dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j - weight[i]] + value[i]);
410+
}
411+
}
403412

404-
return dp[len][size];
413+
return dp[len-1][size]//满足条件的最大值
405414
}
406415

407416
function testWeightBagProblem2 (wight, value, size) {

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