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go版本新解法:双指针逆序遍历,时间复杂度O(n),空间复杂度O(n)

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problems
- 0151.翻转字符串里的单词.md

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`‎problems/0151.翻转字符串里的单词.md‎`

Lines changed: 40 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -639,7 +639,46 @@ func reverse(b []byte) {`
`639`	`639`	`}`
`640`	`640`	`}`
`641`	`641`	```
`642`		`-`
	`642`	+```go
	`643`	`+//双指针解法。指针逆序遍历,将遍历后得到的单词(间隔为空格,用以区分)顺序放置在额外空间`
	`644`	`+//时间复杂度O(n),空间复杂度O(n)`
	`645`	`+func reverseWords(s string) string {`
	`646`	`+ strBytes := []byte(s)`
	`647`	`+ n := len(strBytes)`
	`648`	`+ // 记录有效字符范围的起始和结束位置`
	`649`	`+ start, end := 0, n-1`
	`650`	`+ // 去除开头空格`
	`651`	`+ for start < n && strBytes[start] == 32 {`
	`652`	`+ start++`
	`653`	`+ }`
	`654`	`+ // 处理全是空格或空字符串情况`
	`655`	`+ if start == n {`
	`656`	`+ return ""`
	`657`	`+ }`
	`658`	`+ // 去除结尾空格`
	`659`	`+ for end >= 0 && strBytes[end] == 32 {`
	`660`	`+ end--`
	`661`	`+ }`
	`662`	`+ // 结果切片,预分配容量`
	`663`	`+ res := make([]byte, 0, end-start+1)//这里挺重要的,本人之前没有预分配容量,每次循环都添加单词,导致内存超限(也可能就是我之前的思路有问题)`
	`664`	`+ // 从后往前遍历有效字符范围`
	`665`	`+ for i := end; i >= start; {`
	`666`	`+ // 找单词起始位置,直接通过循环条件判断定位`
	`667`	`+ for ; i >= start && strBytes[i] == 32; i-- {`
	`668`	`+ }`
	`669`	`+ j := i`
	`670`	`+ for ; j >= start && strBytes[j]!= 32; j-- {`
	`671`	`+ }`
	`672`	`+ res = append(res, strBytes[j+1:i+1]...)`
	`673`	`+ // 只在不是最后一个单词时添加空格`
	`674`	`+ if j > start {`
	`675`	`+ res = append(res, 32)`
	`676`	`+ }`
	`677`	`+ i = j`
	`678`	`+ }`
	`679`	`+ return string(res)`
	`680`	`+}`
	`681`	+```
`643`	`682`
`644`	`683`
`645`	`684`	`### JavaScript:`

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`‎problems/0151.翻转字符串里的单词.md‎`

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