【Leetcode】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.

思路：因为题目的条件是第x+1行的第一个数比第x行最后一个数大，所以能够使用二分查找；假设题目条件仅仅说逐行递增，逐列递增，能够从右上角開始查找。此方法也适用于本题。

代码一：

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        const int m = matrix.size();
        if(m == 0)  return false;
        const int n = matrix[0].size();
        if(n == 0)  return false;
        
        int first = 0;
        int last = m * n - 1;
        
        while(first <= last)
        {
            int middle = (first + last) / 2;
            int val = matrix[middle / n][middle % n];
            
            if(val == target)
                return true;
            else if(val < target)
                first = middle + 1;
            else
                last = middle - 1;
        }
        
        return false;
    }
};

代码二：

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        const int m = matrix.size();
        if(m == 0)  return false;
        const int n = matrix[0].size();
        if(n == 0)  return false;
        
        int row = 0;
        int column = n - 1;
        
        while(row < m && column >= 0)
        {
            if(matrix[row][column] == target)
                return true;
            else if(matrix[row][column] > target)
                column--;
            else
                row++;
        }
        
        return false;
    }
};

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