[pull] master from youngyangyang04:master #9

Original file line number	Diff line number	Diff line change
Expand Up		@@ -96,7 +96,7 @@ public:
		vector<vector<int>> merge(vector<vector<int>>& intervals) {
		vector<vector<int>> result;
		if (intervals.size() == 0) return result;
	// 排序的参数使用了lamda表达式
	// 排序的参数使用了lambda表达式
		sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b){return a[0] < b[0];});

		result.push_back(intervals[0]);
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56 changes: 56 additions & 0 deletions problems/0151.翻转字符串里的单词.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -758,7 +758,63 @@ func reverseWord(_ s: inout [Character]) {
		}
		```

	Scala:

	```scala
	object Solution {
	def reverseWords(s: String): String = {
	var sb = removeSpace(s) // 移除多余的空格
	reverseString(sb, 0, sb.length - 1) // 翻转字符串
	reverseEachWord(sb)
	sb.mkString
	}

	// 移除多余的空格
	def removeSpace(s: String): Array[Char] = {
	var start = 0
	var end = s.length - 1
	// 移除字符串前面的空格
	while (start < s.length && s(start) == ' ') start += 1
	// 移除字符串后面的空格
	while (end >= 0 && s(end) == ' ') end -= 1
	var sb = "" // String
	// 当start小于等于end的时候,执行添加操作
	while (start <= end) {
	var c = s(start)
	// 当前字符不等于空,sb的最后一个字符不等于空的时候添加到sb
	if (c != ' ' \|\| sb(sb.length - 1) != ' ') {
	sb ++= c.toString
	}
	start += 1 // 指针向右移动
	}
	sb.toArray
	}

	// 翻转字符串
	def reverseString(s: Array[Char], start: Int, end: Int): Unit = {
	var (left, right) = (start, end)
	while (left < right) {
	var tmp = s(left)
	s(left) = s(right)
	s(right) = tmp
	left += 1
	right -= 1
	}
	}

	// 翻转每个单词
	def reverseEachWord(s: Array[Char]): Unit = {
	var i = 0
	while (i < s.length) {
	var j = i + 1
	// 向后迭代寻找每个单词的坐标
	while (j < s.length && s(j) != ' ') j += 1
	reverseString(s, i, j - 1) // 翻转每个单词
	i = j + 1 // i往后更新
	}
	}
	}
	```



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4 changes: 2 additions & 2 deletions problems/0404.左叶子之和.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -336,8 +336,8 @@ var sumOfLeftLeaves = function(root) {
		if(node===null){
		return 0;
		}
	let leftValue = sumOfLeftLeaves(node.left);
	let rightValue = sumOfLeftLeaves(node.right);
	let leftValue = nodesSum(node.left);
	let rightValue = nodesSum(node.right);
		// 3. 单层递归逻辑
		let midValue = 0;
		if(node.left&&node.left.left===null&&node.left.right===null){
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35 changes: 17 additions & 18 deletions problems/0516.最长回文子序列.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -186,29 +186,28 @@ class Solution:
		Go:
		```Go
		func longestPalindromeSubseq(s string) int {
	lenth:=len(s)
	dp:=make([][]int,lenth)
	for i:=0;i<lenth;i++{
	for j:=0;j<lenth;j++{
	if dp[i]==nil{
	dp[i]=make([]int,lenth)
	}
	if i==j{
	dp[i][j]=1
	}
	size := len(s)
	max := func(a, b int) int {
	if a > b {
	return a
		}
	return b
		}
	for i:=lenth-1;i>=0;i--{
	for j:=i+1;j<lenth;j++{
	if s[i]==s[j]{
	dp[i][j]=dp[i+1][j-1]+2
	}else {
	dp[i][j]=max(dp[i+1][j],dp[i][j-1])
	dp := make([][]int, size)
	for i := 0; i < size; i++ {
	dp[i] = make([]int, size)
	dp[i][i] = 1
	}
	for i := size - 1; i >= 0; i-- {
	for j := i + 1; j < size; j++ {
	if s[i] == s[j] {
	dp[i][j] = dp[i+1][j-1] + 2
	} else {
	dp[i][j] = max(dp[i][j-1], dp[i+1][j])
		}
		}
		}

	return dp[0][lenth-1]
	return dp[0][size-1]
		}
		```

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3 changes: 2 additions & 1 deletion problems/0739.每日温度.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -34,7 +34,8 @@

		那么单调栈的原理是什么呢?为什么时间复杂度是O(n)就可以找到每一个元素的右边第一个比它大的元素位置呢?

	单调栈的本质是空间换时间,因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素大的元素,优点是只需要遍历一次。

	单调栈的本质是空间换时间,因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素高的元素,优点是只需要遍历一次。


		在使用单调栈的时候首先要明确如下几点:
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29 changes: 28 additions & 1 deletion problems/剑指Offer58-II.左旋转字符串.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -290,7 +290,34 @@ func reverseString(_ s: inout [Character], startIndex: Int, endIndex: Int) {
		}
		```


	Scala:

	```scala
	object Solution {
	def reverseLeftWords(s: String, n: Int): String = {
	var str = s.toCharArray // 转换为Array
	// abcdefg => ba cdefg
	reverseString(str, 0, n - 1)
	// ba cdefg => ba gfedc
	reverseString(str, n, str.length - 1)
	// ba gfedc => cdefgab
	reverseString(str, 0, str.length - 1)
	// 最终返回,return关键字可以省略
	new String(str)
	}
	// 翻转字符串
	def reverseString(s: Array[Char], start: Int, end: Int): Unit = {
	var (left, right) = (start, end)
	while (left < right) {
	var tmp = s(left)
	s(left) = s(right)
	s(right) = tmp
	left += 1
	right -= 1
	}
	}
	}
	```



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