Conquering LeetCode #12

Easy question: 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.

This is probably the easiest problem I’ve ever solved. The only somewhat tricky part was realizing that a word could end with a space. The solution was to trim the string before doing anything with it.

public class Solution {
public int lengthOfLastWord(String s) {
int counter = 0;
s = s.trim();
for (int i = s.length() - 1; i >= 0; i--){
if (s.charAt(i) == ' '){
return counter;
return s.length();

— -

The second problem I solved was just as easy: Plus One

Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
You may assume the integer do not contain any leading zero, except the number 0 itself.
The digits are stored such that the most significant digit is at the head of the list.

At first glance, its simple. Just add a 1 to the end of the array. But if the end of the array is a 9, then set that to 0 and add 1 to the second-last element, and so on.

When I submitted my solution, it failed a test case where digits was [9]. The result was supposed to be [1, 0]. I had incorrectly assumed that the head of the array would never be 9. Whoops. Shouldn’t assume things.

But anyway, creating a new array wasn’t a problem. It was even easier when I remembered that initializing an integer array in Java sets all values to be 0. So I just had to change the first value to a 1.

public class Solution {
public int[] plusOne(int[] digits) {
for (int i = digits.length - 1; i >= 0; i--){
if (digits[i] == 9){
digits[i] = 0;
} else {
digits[i] = digits[i] + 1;
return digits;

int[] toReturn = new int[digits.length + 1];
toReturn[0] = 1;
return toReturn;
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