[LeetCode]Length of Last Word&Plus One

发现还早，并没有睡意，再看了一眼后面的Easy，发现很快，so，写了吧，为了节约能源，把两题合在一起

---

Length of Last Word

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.

• 题意：给一个串，求最后一个单词的长度.
• 解法：从后往前循环，从第一个非空格开始计数，到空格或者循环完结束.

class Solution {public:    int lengthOfLastWord(string s) {        int iLastChar = s.size() - 1;        for (; iLastChar >= 0 && ' ' == s[iLastChar]; --iLastChar);        int iResult = 0;        for (; iLastChar >= 0 && ' ' != s[iLastChar]; --iLastChar, ++iResult);        return iResult;    }};

Plus One

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

• 题意：大数加1..
• 解法：写过大数的这个应该没啥问题，需要注意的地方只有一点，进位，比如999，要在前面insert一个1.

class Solution {public:    vector<int> plusOne(vector<int>& digits) {        int iAddValue = 1;        for (int i = digits.size() - 1; i >= 0 && iAddValue > 0; --i) {            digits[i] += iAddValue;            iAddValue = digits[i] > 9;            digits[i] -= iAddValue ? 10 : 0;        }        if (iAddValue)            digits.insert(digits.begin(), 1);        return digits;    }};

欢迎访问我的github,我的leetcode持续更新: https://github.com/tsfissure/LeetCode