[C++]LeetCode: 61 Search a 2D Matrix

题目：

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target =

3

, return

true

.

思路：根据题意，矩阵每行都有序排列，并且逐行增加，所以我们可以把MN矩阵展开成一维有序数组。

num表示元素的序列号，从0开始。i, j表示矩阵中的行列坐标，从0开始。

Num = n*i + j; => i = Num /n ; j = Num % n; n表示矩阵的列。（画图可以得到）

接下来利用二分查找，找到target.

Attention: 注意将矩阵中的序列号和矩阵行列号的转换。

int mid = lo + (hi - lo) /2;
 int i = mid / n;
 int j = mid % n;

复杂度：O(log(N))

AC Code: （MY_CODE ONE_PASS）

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        //其实类似于将mn矩阵展开成一个数组然后查找，不过需要根据第几个元素，计算行列
        //num表示元素的序列号，从0开始。i,j表示矩阵中的行列坐标，从0开始。
        //Num = n*i + j;  => i = Num /n ; j = Num % n; n表示矩阵的列。
        
        bool ret = false;
        if(matrix.size() == 0 || matrix[0].size() == 0) return ret;
        int m = matrix.size();
        int n = matrix[0].size();
        
        int lo = 0;
        int hi = (m - 1) * n + (n - 1);  // hi = m * n -1;
        
        while(lo <= hi)
        {
            int mid = lo + (hi - lo) /2;
            int i = mid / n;
            int j = mid % n;
            if(target > matrix[i][j])
            {
                lo = mid + 1;
            }
            else if(target < matrix[i][j])
            {
                hi = mid - 1;
            }
            else
            {
               ret = true;
               break;
            }
        }
        
        return ret;
    
    }
};