Commit defa303

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Create Length of Last Word

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	`1`	`+想法:由後往前找,看到的第一個非空白的字母就開始紀錄往前連續幾個並回傳`
	`2`	`+如果都沒有找到代表字串全部都是空白,回傳 0`
	`3`	`+`
	`4`	`+Time Complexity : O(n) for traversing the string`
	`5`	`+Space Complexity : O(1) for variables`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ int lengthOfLastWord(string s) {`
	`10`	`+ for(int i = s.length() - 1 ; i >= 0 ; i--) {`
	`11`	`+ if ( s[i] == ' ' )`
	`12`	`+ continue ;`
	`13`	`+ int count = 0 ;`
	`14`	`+ while ( i >= 0 && s[i] != ' ') {`
	`15`	`+ count++ ;`
	`16`	`+ i-- ;`
	`17`	`+ }`
	`18`	`+ return count ;`
	`19`	`+ }`
	`20`	`+ return 0 ;`
	`21`	`+ }`
	`22`	`+};`

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