题目描述(Medium)

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.

题目链接

https://leetcode.com/problems/search-a-2d-matrix/description/

Example 1:

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

Example 2:

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

算法分析

方法一：首先选取右上角数字，如果该数字等于要查找的数字，查找过程结束；如果该数字大于要查找的数字，去掉此数字所在列；如果该数字小于要查找的数字，则去掉该数字所在行。重复上述过程直到找到要查找的数字，或者查找范围为空。

方法二：二分法查找。

提交代码(方法一)：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.size() < 1 || matrix[0].size() < 1) return false;
        int rows = matrix.size(), cols = matrix[0].size();
        int row = 0, col = cols - 1;
            
        while (row < rows && col >= 0)
        {
            if (matrix[row][col] < target)
                ++row;
            else if (matrix[row][col] > target)
            {
                // 去掉这一列及下面的行
                --col;
                rows = row + 1;
            }
            else
                return true;
        }
        
        return false;
    }
};

提交代码(方法二)：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.size() < 1 || matrix[0].size() < 1) return false;
        
        const size_t m = matrix.size();
        const size_t n = matrix[0].size();
        int first = 0, last = m * n - 1;
        
        while (first <= last)
        {
            int mid = (first + last) >> 1;
            int value = matrix[mid / n][mid % n];
            
            if (value == target)
                return true;
            else if (value < target)
                first = mid + 1;
            else
                last = mid - 1;
        }
        return false;
    }
};