[LeetCode] 58. Length of Last Word

轉自LeetCode

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example,
Given s = "Hello World",
return 5.

<Solution>

不難，就是字串的處理

先看一下有哪些情況

"hello"
""
"hello world"
"a "
" a"
"     "

根據測資，解題想法如下

• 因為是要看最後一個 word，所以從後面往前找

• 遇到空白就略過，如果都沒遇到字元，就回傳0

• 找到字元後，就往前找到空白直到字元用完

code 如下

	class Solution {
	public:
	int lengthOfLastWord(string s) {
	if(s.empty()) {
	return 0;
	}
	int pos = s.length() - 1;
	//> find alphabet
	while(pos >= 0 && s[pos] == ' ') {
	--pos;
	}
	if(pos < 0) {
	return 0;
	}
	int right = pos;
	//> find last word
	while(pos >= 0 && s[pos] != ' ') {
	--pos;
	}
	return right - pos;
	}
	};

view raw lengthOfLastWord.cpp hosted with ❤ by GitHub