[LeetCode] Search a 2D Matrix

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 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

从左上角到右下角的顺序遍历数组，如果左上角元素大于target，则左移一位，直到元素小于target时，开始从该列从上到下遍历，直到找到target。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.empty())
            return false;
        int m = matrix.size();
        int n = matrix[0].size();
        int i = 0, j = n-1;
        while (i < m && j >= 0)
        {
            if (matrix[i][j] == target)
                return true;
            else if (matrix[i][j] > target)
                --j;
            else if (matrix[i][j] < target)
                ++i;
        }
        return false;
    }
};