240. Search a 2D Matrix II

解法一

思路

I found that if we start from the left bottom cornor of the matrix, the right neighbor is always larger than the element, and the top neighbor is always less than the element. So we can set two pointers one move in column and the other move in row to find the target.
Basic idea is, while both pointers are not out of bound, we compare the target with the pivot element. If target is larger we move one step to the right, if smaller, we move one step to the top. If target is equal to the element, we return true.

程式碼

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int rows = matrix.length;
        if (rows == 0 || martix == null) {
            return false;
        }
        int cols = matrix[0].length;
        if (matrix[0].length == 0 || matrix == null) {
            return false;
        }

        int i = rows - 1;
        int j = 0;
        while (i >= 0 && j < cols) {
            if (target > matrix[i][j]) {
                j++;
            } else if (target < matrix[i][j]) {
                i--;
            } else {
                return true;
            }
        }
        return false;
    }
}

複雜度分析

• 時間複雜度
  The worst case would be when we start at the left bottom cornor and end at right top cornor. O(m + n).
• 空間複雜度
  O(1)

解法二

思路

程式碼

複雜度分析

時間複雜度

• 最好情況
• 最壞情況
• 平均情況
  空間複雜度

Takeaway