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.

方法一：每行中，从左到右元素依次增大，每列中，从上到下元素依次增大，因此可以从右上角开始查找，若target小于matrix[i][j]，那么--j，若target大于matrix[i][j]，那么++i，继续比较target与matrix[i][j]，直至找到或找不到

时间复杂度为O(m+n)

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rows = matrix.size();
        int cols = matrix[0].size();
        int i = 0, j = cols-1;
        while( i<rows && j>=0 ) {
            if( matrix[i][j] == target ) return true;
            if( target < matrix[i][j] ) --j;    //target小于matrix[i][j]，那么往左边找
            else ++i;   //target大于matrix[i][j]，那么往下面找
        }
        return false;
    }
};

方法二：The first integer of each row is greater than the last integer of the previous row.

由于每行的第一个元素都比前面行的最后一个元素都大，因此可以考虑二分搜索

现在第一列二分搜索target，若找到，则直接返回true，找不到则返回target应该在的那一行，然后继续在那一行继续二分搜索target，时间复杂度为O(logm+logn)

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rows = matrix.size();
        int cols = matrix[0].size();
        int low = 0, high = rows-1;
        while( low <= high ) {  //在第一列进行二分搜索
            int mid = low + (high - low)/2;
            if( matrix[mid][0] == target ) return true;
            if( target < matrix[mid][0] ) high = mid-1;
            else {
                if( target > matrix[mid][cols-1] )  //若大于第一个元素及最后一个元素
                    low = mid+1;
                else {  //若小于最后一个元素，那么target理论上应该在mid所在的行中
                    low = mid;
                    break;
                }
            }
        }
        if( low >= rows ) return false;
        int row = low;
        low = 0, high = cols-1;
        while( low <= high ) {  //在搜到的行进行二分搜索
            int mid = low + (high-low)/2;
            if( target == matrix[row][mid] ) return true;
            if( target > matrix[row][mid] ) low = mid+1;
            else high = mid-1;
        }
        return false;
    }
};