[pull] master from youngyangyang04:master #13

Original file line number	Diff line number	Diff line change
Expand Up		@@ -186,6 +186,24 @@ const maxSubArray = nums => {
		};
		```

	TypeScript:

	```typescript
	function maxSubArray(nums: number[]): number {
	/**
	dp[i]:以nums[i]结尾的最大和
	*/
	const dp: number[] = []
	dp[0] = nums[0];
	let resMax: number = 0;
	for (let i = 1; i < nums.length; i++) {
	dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);
	resMax = Math.max(resMax, dp[i]);
	}
	return resMax;
	};
	```



		-----------------------
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37 changes: 37 additions & 0 deletions problems/0072.编辑距离.md

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Expand Up		@@ -327,5 +327,42 @@ const minDistance = (word1, word2) => {
		};
		```

	TypeScript:

	```typescript
	function minDistance(word1: string, word2: string): number {
	/**
	dp[i][j]: word1前i个字符,word2前j个字符,最少操作数
	dp[0][0]=0:表示word1前0个字符为'', word2前0个字符为''
	*/
	const length1: number = word1.length,
	length2: number = word2.length;
	const dp: number[][] = new Array(length1 + 1).fill(0)
	.map(_ => new Array(length2 + 1).fill(0));
	for (let i = 0; i <= length1; i++) {
	dp[i][0] = i;
	}
	for (let i = 0; i <= length2; i++) {
	dp[0][i] = i;
	}
	for (let i = 1; i <= length1; i++) {
	for (let j = 1; j <= length2; j++) {
	if (word1[i - 1] === word2[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1];
	} else {
	dp[i][j] = Math.min(
	dp[i - 1][j],
	dp[i][j - 1],
	dp[i - 1][j - 1]
	) + 1;
	}
	}
	}
	return dp[length1][length2];
	};
	```



		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

20 changes: 20 additions & 0 deletions problems/0101.对称二叉树.md

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Expand Up		@@ -725,5 +725,25 @@ func isSymmetric3(_ root: TreeNode?) -> Bool {
		}
		```

	## Scala

	递归:
	```scala
	object Solution {
	def isSymmetric(root: TreeNode): Boolean = {
	if (root == null) return true // 如果等于空直接返回true
	def compare(left: TreeNode, right: TreeNode): Boolean = {
	if (left == null && right == null) return true // 如果左右都为空,则为true
	if (left == null && right != null) return false // 如果左空右不空,不对称,返回false
	if (left != null && right == null) return false // 如果左不空右空,不对称,返回false
	// 如果左右的值相等,并且往下递归
	left.value == right.value && compare(left.left, right.right) && compare(left.right, right.left)
	}
	// 分别比较左子树和右子树
	compare(root.left, root.right)
	}
	}
	```

		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

30 changes: 30 additions & 0 deletions problems/0115.不同的子序列.md

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Expand Up		@@ -267,6 +267,36 @@ const numDistinct = (s, t) => {
		};
		```

	TypeScript:

	```typescript
	function numDistinct(s: string, t: string): number {
	/**
	dp[i][j]: s前i个字符,t前j个字符,s子序列中t出现的个数
	dp[0][0]=1, 表示s前0个字符为'',t前0个字符为''
	*/
	const sLen: number = s.length,
	tLen: number = t.length;
	const dp: number[][] = new Array(sLen + 1).fill(0)
	.map(_ => new Array(tLen + 1).fill(0));
	for (let m = 0; m < sLen; m++) {
	dp[m][0] = 1;
	}
	for (let i = 1; i <= sLen; i++) {
	for (let j = 1; j <= tLen; j++) {
	if (s[i - 1] === t[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
	} else {
	dp[i][j] = dp[i - 1][j];
	}
	}
	}
	return dp[sLen][tLen];
	};
	```




		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

48 changes: 48 additions & 0 deletions problems/0226.翻转二叉树.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -818,5 +818,53 @@ func invertTree(_ root: TreeNode?) -> TreeNode? {
		}
		```

	### Scala

	深度优先遍历(前序遍历):
	```scala
	object Solution {
	def invertTree(root: TreeNode): TreeNode = {
	if (root == null) return root
	// 递归
	def process(node: TreeNode): Unit = {
	if (node == null) return
	// 翻转节点
	val curNode = node.left
	node.left = node.right
	node.right = curNode
	process(node.left)
	process(node.right)
	}
	process(root)
	root
	}
	}
	```

	广度优先遍历(层序遍历):
	```scala
	object Solution {
	import scala.collection.mutable
	def invertTree(root: TreeNode): TreeNode = {
	if (root == null) return root
	val queue = mutable.Queue[TreeNode]()
	queue.enqueue(root)
	while (!queue.isEmpty) {
	val len = queue.size
	for (i <- 0 until len) {
	var curNode = queue.dequeue()
	if (curNode.left != null) queue.enqueue(curNode.left)
	if (curNode.right != null) queue.enqueue(curNode.right)
	// 翻转
	var tmpNode = curNode.left
	curNode.left = curNode.right
	curNode.right = tmpNode
	}
	}
	root
	}
	}
	```

		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

25 changes: 25 additions & 0 deletions problems/0392.判断子序列.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -201,7 +201,32 @@ const isSubsequence = (s, t) => {
		};
		```

	TypeScript:

	```typescript
	function isSubsequence(s: string, t: string): boolean {
	/**
	dp[i][j]: s的前i-1个,t的前j-1个,最长公共子序列的长度
	*/
	const sLen: number = s.length,
	tLen: number = t.length;
	const dp: number[][] = new Array(sLen + 1).fill(0)
	.map(_ => new Array(tLen + 1).fill(0));
	for (let i = 1; i <= sLen; i++) {
	for (let j = 1; j <= tLen; j++) {
	if (s[i - 1] === t[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1] + 1;
	} else {
	dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
	}
	}
	}
	return dp[sLen][tLen] === s.length;
	};
	```

		Go:

		```go
		func isSubsequence(s string, t string) bool {
		dp := make([][]int,len(s)+1)
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61 changes: 61 additions & 0 deletions problems/0583.两个字符串的删除操作.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -229,6 +229,67 @@ const minDistance = (word1, word2) => {
		};
		```

	TypeScript:

	> dp版本一:

	```typescript
	function minDistance(word1: string, word2: string): number {
	/**
	dp[i][j]: word1前i个字符,word2前j个字符,所需最小步数
	dp[0][0]=0: word1前0个字符为'', word2前0个字符为''
	*/
	const length1: number = word1.length,
	length2: number = word2.length;
	const dp: number[][] = new Array(length1 + 1).fill(0)
	.map(_ => new Array(length2 + 1).fill(0));
	for (let i = 0; i <= length1; i++) {
	dp[i][0] = i;
	}
	for (let i = 0; i <= length2; i++) {
	dp[0][i] = i;
	}
	for (let i = 1; i <= length1; i++) {
	for (let j = 1; j <= length2; j++) {
	if (word1[i - 1] === word2[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1];
	} else {
	dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
	}
	}
	}
	return dp[length1][length2];
	};
	```

	> dp版本二:

	```typescript
	function minDistance(word1: string, word2: string): number {
	/**
	dp[i][j]: word1前i个字符,word2前j个字符,最长公共子序列的长度
	dp[0][0]=0: word1前0个字符为'', word2前0个字符为''
	*/
	const length1: number = word1.length,
	length2: number = word2.length;
	const dp: number[][] = new Array(length1 + 1).fill(0)
	.map(_ => new Array(length2 + 1).fill(0));
	for (let i = 1; i <= length1; i++) {
	for (let j = 1; j <= length2; j++) {
	if (word1[i - 1] === word2[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1] + 1;
	} else {
	dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
	}
	}
	}
	const maxLen: number = dp[length1][length2];
	return length1 + length2 - maxLen * 2;
	};
	```




		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

65 changes: 65 additions & 0 deletions problems/0763.划分字母区间.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -158,6 +158,71 @@ class Solution {
		return list;
		}
		}

	class Solution{
	/解法二: 上述c++补充思路的Java代码实现/

	public int[][] findPartitions(String s) {
	List<Integer> temp = new ArrayList<>();
	int[][] hash = new int[26][2];//26个字母2列表示该字母对应的区间

	for (int i = 0; i < s.length(); i++) {
	//更新字符c对应的位置i
	char c = s.charAt(i);
	if (hash[c - 'a'][0] == 0) hash[c - 'a'][0] = i;

	hash[c - 'a'][1] = i;

	//第一个元素区别对待一下
	hash[s.charAt(0) - 'a'][0] = 0;
	}


	List<List<Integer>> h = new LinkedList<>();
	//组装区间
	for (int i = 0; i < 26; i++) {
	//if (hash[i][0] != hash[i][1]) {
	temp.clear();
	temp.add(hash[i][0]);
	temp.add(hash[i][1]);
	//System.out.println(temp);
	h.add(new ArrayList<>(temp));
	// }
	}
	// System.out.println(h);
	// System.out.println(h.size());
	int[][] res = new int[h.size()][2];
	for (int i = 0; i < h.size(); i++) {
	List<Integer> list = h.get(i);
	res[i][0] = list.get(0);
	res[i][1] = list.get(1);
	}

	return res;

	}

	public List<Integer> partitionLabels(String s) {
	int[][] partitions = findPartitions(s);
	List<Integer> res = new ArrayList<>();
	Arrays.sort(partitions, (o1, o2) -> Integer.compare(o1[0], o2[0]));
	int right = partitions[0][1];
	int left = 0;
	for (int i = 0; i < partitions.length; i++) {
	if (partitions[i][0] > right) {
	//左边界大于右边界即可纪委一次分割
	res.add(right - left + 1);
	left = partitions[i][0];
	}
	right = Math.max(right, partitions[i][1]);

	}
	//最右端
	res.add(right - left + 1);
	return res;

	}
	}
		```

		### Python
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24 changes: 24 additions & 0 deletions problems/1035.不相交的线.md

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Expand Up		@@ -183,6 +183,30 @@ const maxUncrossedLines = (nums1, nums2) => {
		};
		```

	TypeScript:

	```typescript
	function maxUncrossedLines(nums1: number[], nums2: number[]): number {
	/**
	dp[i][j]: nums1前i-1个,nums2前j-1个,最大连线数
	*/
	const length1: number = nums1.length,
	length2: number = nums2.length;
	const dp: number[][] = new Array(length1 + 1).fill(0)
	.map(_ => new Array(length2 + 1).fill(0));
	for (let i = 1; i <= length1; i++) {
	for (let j = 1; j <= length2; j++) {
	if (nums1[i - 1] === nums2[j - 1]) {
	dp[i][j] = dp[i - 1][j - 1] + 1;
	} else {
	dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
	}
	}
	}
	return dp[length1][length2];
	};
	```




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