Oct 11, 2022 · Oct 2, 2022 · Oct 3, 2022 · Oct 3, 2022 · Oct 3, 2022 · Oct 4, 2022
diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md
Original file line number	Diff line number	Diff line change
Expand Up		@@ -360,6 +360,74 @@ class Solution {
		}
		```

	```java
	/*
	* 解法四:时间复杂度 O(n)
	* 参考卡哥 c++ 代码的实现:先移除多余空格,再将整个字符串反转,最后把单词逐个反转
	*/
	class Solution {
	//用 char[] 来实现 String 的 removeExtraSpaces,reverse 操作
	public String reverseWords(String s) {
	char[] chars = s.toCharArray();
	//1.去除首尾以及中间多余空格
	chars = removeExtraSpaces(chars);
	//2.整个字符串反转
	reverse(chars, 0, chars.length - 1);
	//3.单词反转
	reverseEachWord(chars);
	return new String(chars);
	}

	//1.用快慢指针去除首尾以及中间多余空格,可参考数组元素移除的题解
	public char[] removeExtraSpaces(char[] chars) {
	int slow = 0;
	for (int fast = 0; fast < chars.length; fast++) {
	//先用 fast 移除所有空格
	if (chars[fast] != ' ') {
	//在用 slow 加空格。除第一个单词外,单词末尾要加空格
	if (slow != 0)
	chars[slow++] = ' ';
	//fast 遇到空格或遍历到字符串末尾,就证明遍历完一个单词了
	while (fast < chars.length && chars[fast] != ' ')
	chars[slow++] = chars[fast++];
	}
	}
	//相当于 c++ 里的 resize()
	char[] newChars = new char[slow];
	System.arraycopy(chars, 0, newChars, 0, slow);
	return newChars;
	}

	//双指针实现指定范围内字符串反转,可参考字符串反转题解
	public void reverse(char[] chars, int left, int right) {
	if (right >= chars.length) {
	System.out.println("set a wrong right");
	return;
	}
	while (left < right) {
	chars[left] ^= chars[right];
	chars[right] ^= chars[left];
	chars[left] ^= chars[right];
	left++;
	right--;
	}
	}

	//3.单词反转
	public void reverseEachWord(char[] chars) {
	int start = 0;
	//end <= s.length() 这里的 = ,是为了让 end 永远指向单词末尾后一个位置,这样 reverse 的实参更好设置
	for (int end = 0; end <= chars.length; end++) {
	// end 每次到单词末尾后的空格或串尾,开始反转单词
	if (end == chars.length \|\| chars[end] == ' ') {
	reverse(chars, start, end - 1);
	start = end + 1;
	}
	}
	}
	}
	```

		python:

		```Python
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