【Search a 2D Matrix】cpp

题目：

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.

代码：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
            if (matrix.size()==0) return false;
            const int ROW = matrix.size();
            const int COL = matrix[0].size();
            // search the target row
            int begin = 0;
            int end = ROW-1;
            while ( begin<=end )
            {
                int mid = (begin+end)/2;
                int lower = matrix[mid][0];
                int upper = matrix[mid][COL-1];
                if ( target>=lower && target<=upper )
                {
                    return Solution::binarySearch(matrix[mid], target);
                }
                if ( target<lower )
                {
                    end = mid-1;
                    continue;
                }
                if ( target>upper )
                {
                    begin = mid+1;
                    continue;
                }
            }
            return false;
    }
    static bool binarySearch(vector<int>& row, int target)
    {
            int begin = 0;
            int end = row.size()-1;
            while ( begin<=end )
            {
                int mid = (begin+end)/2;
                if ( row[mid]==target ) return true;
                if ( row[mid]>target )
                {
                    end = mid-1;
                }
                else
                {
                    begin = mid+1;
                }
            }
            return false;
    }
};

tips：

1. 首先二分查找可能所在的行

2. 确定某一行之后，再二分查找所在的列

完毕。

======================================

另一种思路：把大的二维矩阵当成一个一维数组看，只需要执行一次二分查找就OK了。

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

tips：

注意这条语句“int val = matrix[mid/COL][mid%COL]”。

一开始写成了"int val = matrix[mid/ROW][mid%COL]" 掉入了这个思维陷阱，因为每行有多少列应该是基础长度单元，所以不论是取商还是余数，分母上都应该是COL。

============================================

第二次过这道题，沿用一般的思路，先找可能在哪一行；再去行里面找。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
            // search for row
            int row = -1;
            int begin = 0;
            int end = matrix.size()-1;
            while ( begin<=end )
            {
                int mid = (begin+end)/2;
                if ( matrix[mid][0]<=target && matrix[mid][matrix[mid].size()-1]>=target )
                {
                    row = mid;
                    break;
                }
                else if ( matrix[mid][0]>target )
                {
                    end = mid-1;
                }
                else
                {
                    begin = mid+1;
                }
            }
            if ( row==-1 ) return false;
            // search in the row
            begin = 0;
            end = matrix[row].size()-1;
            while ( begin<=end )
            {
                int mid = (begin+end)/2;
                if ( matrix[row][mid]==target ) return true;
                if ( matrix[row][mid]>target )
                {
                    end = mid-1;
                }
                else
                {
                    begin = mid+1;
                }
            }
            return false;
    }
};