LeetCode 74. 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.
  For example,

Consider the following matrix:

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

Given target = 3, return true.

根据矩阵的性质可知

• 先判断该target可能所在的行
• 然后在判断改行中是否出现该值
• 两次判断都可以用二分

所以复杂度为O(log(n))
代码如下：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() == 0)
            return false;

        int b = 0, e = matrix.size()-1;
        while(b < e)
        {
            int mid = b + (e - b + 1) / 2;
            if(matrix[mid][0] == target)
                return true;
            else if(matrix[mid][0] < target)
                b = mid;
            else
                e = mid - 1;
        }
        int x = 0, y = matrix[b].size() - 1;
        while(x < y)
        {
            int mid = x + (y - x + 1) / 2;
            if(matrix[b][mid] == target)
                return true;
            else if(matrix[b][mid] < target)
                x = mid;
            else
                y = mid - 1;
        }
        if(b < matrix.size() && x < matrix[b].size() && matrix[b][x] == target)
            return true;
        return false;
    }
};