LeetCode | 74. Search a 2D Matrix

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#leetcode #matrix #搜索 #算法

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本文介绍了一种高效的矩阵搜索算法，该算法能在mxn的矩阵中寻找特定值，此矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了两种实现方式，一种是对行和列分别进行二分查找，另一种是将矩阵视为一维数组进行查找。

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：二分，复杂度O(log(m+n))

Solution

代码一：先对行二分，再对列二分

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int row = -1, col = -1;
        int m = matrix.size();

        if(m == 0)
            return false;

        int n = matrix[0].size();

        if(n == 0)
            return false;

        //先找在哪一行
        int left = 0, right = m-1;
        while(left <= right)
        {
            int mid = (left+right)/2;
            if(matrix[mid][0] > target)
                right = mid-1;
            else
            {
                if(mid == m-1)  //已经是最后一行
                {
                    row = mid; break;
                }
                else if(matrix[mid+1][0] > target)
                {
                    row = mid; break;
                }
                else
                    left = mid+1;
            }
        }
        if(row == -1)
            return false;
        left = 0; right = n-1;
        while(left <= right)
        {
            int mid = (left+right)/2;
            if(matrix[row][mid] == target)
            {
                col = mid; break;
            }
            else if(matrix[row][mid] > target)
                right = mid-1;
            else
                left = mid+1;
        }

        if(col == -1)
            return false;
        return true;
    }
};

代码二：将矩阵展开，当做一个有序列表，然后直接查找

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

        int l = 0, r = m * n - 1;
        while (l != r){
            int mid = (l + r - 1) >> 1;
            if (matrix[mid / m][mid % m] < target)
                l = mid + 1;
            else 
                r = mid;
        }
        return matrix[r / m][r % m] == target;
    }
};