Search a 2D Matrix II(观察矩阵规律)

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        // 从一个角落开始搜eg: 左下角 或者右上角
        // 以左下角开始为例： 比当前小则向上搜索 比当前大则向左搜索 最大搜索 n+m步  最少搜索min(m,n)
        int imax=matrix.size(), jmax=matrix[0].size(), i=imax-1, j=0; // 行 列 总数
        // 从左下角开始
        while(i >= 0 && j<jmax){
            int tmp= matrix[i][j];
            if(target > tmp) j++;
            else if(target < tmp) i--;
            else return true;
        }
        return false;
    }
};