学步园

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：这是一道二分搜索题。先对行二分搜索，再对列二分搜索。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty())
        {
            return false;
        }
        int rowL = 0, rowR = matrix.size() - 1, rowM;
        int colL = 0, colR = matrix[0].size() - 1, colM;
        while (rowL <= rowR) {
            rowM = (rowL + rowR) / 2;
            if (target < matrix[rowM][colL]) {
                rowR = rowM - 1;
            }
            else if(target > matrix[rowM][colR]) {
                rowL = rowM + 1;
            }
            else {
                break;
            }
        }
        while(colL <= colR) {
            colM = (colL + colR) / 2;
            if (target < matrix[rowM][colM]) {
                colR = colM - 1;
            }
            else if(target > matrix[rowM][colM]) {
                colL = colM + 1;
            }
            else {
                return true;
            }
        }
        return false;
    }
};