Question

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

  • Integers in each row are sorted from left to right.
  • The first integer of each row is greater than the last integer of the previous row.
Example
Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Thinking:

  • Method1:从右上角开始遍历
    • 如果temp大于target，左移。
    • 如果temp小于target，下移。
    • 如果等于，返回。
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0) return false;
        int height = matrix.length;
        int width = matrix[0].length;
        int row = 0, col = width - 1;
        while(row < height && col >= 0){
            int temp = matrix[row][col];
            if(temp == target) return true;
            else if(temp < target)
                row++;
            else
                col--;
        }
        return false;
    }
}

二刷

这道题还是比较经典，二刷的完全可以记住一刷的结果。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int height = matrix.length, width = matrix[0].length;
        int col = width - 1, row = 0;
        while(row < height && col >= 0){
            if(matrix[row][col] == target) return true;
            else if(matrix[row][col] > target) col--;
            else row++;
        }
        return false;
    }
}