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.

10/3/2015 update

//[start, end] inclusive, it may stop at 1 element left or 2, 
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int start = 0;
        int end = matrix.size() - 1;
        while(end - start > 1){
            int mid = (start + end) / 2;
            if(matrix[mid][0] > target){
                end = mid - 1;
            }
            else{
                start = mid;
            }
        }
        if(end - start == 1){
            start = matrix[end][0] <= target?end:start;//BUG HERE, <= not <
        }
        int row = start;
        start = 0;
        end = matrix[0].size() - 1;
        while(end >= start){
            int mid = (start + end) / 2;
            if(matrix[row][mid] == target){
                return true;
            }
            else if(matrix[row][mid] > target){
                end = mid - 1;
            }
            else{
                start = mid + 1;
            }
        }
        return false;
        
    }
};

两次二分查找，第一次找可能的行，第二次确定列。

class Solution {
public:
    //apply two binary search process
    //first search the possible row
    //second search the exact column
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty()) return false;
        size_t height = matrix.size();
        size_t width = matrix.front().size();
        int row = findRow(matrix, 0, height, target);
        if(row == -1) return false;
        int column = findColumn(matrix[row], 0, width, target);
        return column != -1;
    }
    int findRow(vector<vector<int>> &matrix, int start, int end, int target){
        //base case
        if(end - start == 1){
            if(matrix[start][0] <= target){
                return start;
            }
            else{
                return -1;
            }
        }
        int mid = (start + end) / 2;
        if(matrix[mid][0] <= target){
            return findRow(matrix, mid, end, target);
        }
        else{
            return findRow(matrix, start, mid, target);
        }
    }
    int findColumn(vector<int> &row, int start, int end, int target){
        //base case
        if(end - start == 1){
            if(row[start] == target){
                return start;
            }
            else{
                return -1;
            }
        }
        int mid = (start + end) / 2;
        if(row[mid] <= target){
            return findColumn(row, mid, end, target);
        }
        else{
            return findColumn(row, start, mid, target);
        }
    }
};