readme.md

Leetcode

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word (last word means the last appearing word if we loop from left to right) in the string.

If the last word does not exist, return 0.

Note: A word is defined as a maximal substring consisting of non-space characters only.

Example:

Input: "Hello World" Output: 5

Solution

package leetcode59_SpiralMatrixII;

class Solution {
  private int[][] result;
  private int index = 1;
  private int[][] directions = new int[][] {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

  public int[][] generateMatrix(int n) {
    if (n == 0) return new int[][] {};
    if (n == 1) return new int[][] {{1}};

    result = new int[n][n];
    for (int i = 0; i < (n + 1) / 2; i++) {
      result[i][i] = index;
      index++;
      goCircle(i, i);
    }

    return result;
  }

  private void goCircle(int x, int y) {
    int newX = x, newY = y;
    int length = result.length;

    for (int i = 0; i < directions.length; i++) {
      while (true) {
        newX += directions[i][0];
        newY += directions[i][1];

        if (newX >= 0 && newY >= 0 && newX < length && newY < length && result[newX][newY] == 0) {
          result[newX][newY] = index;
          index++;
        } else {
          newX -= directions[i][0];
          newY -= directions[i][1];
          break;
        }
      }
    }
  }
}