LeetCode之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.

我的解法：对列应用一次二分查找，对选中的行再一次二分查找。

class Solution {
public:
   bool searchMatrix(vector<vector<int> > &matrix, int target) {
		 int rownum = matrix.size();
		 int colnum = matrix[0].size();
         
         int len=rownum;
         int half,mid,first=0;
         while(len>0){
             half=len>>1;
             mid=first+half;
             if(matrix[mid][colnum-1]<target){
                 first=mid+1;
                 len=len-half-1;
             }
             else
                len=half;
         }//while
         if(first==rownum) return false;
		 return binarySearch(matrix[first], colnum, target);
		 
	 }

	
	 bool binarySearch(vector<int> A, int length, int target){
		 int left = 0, right = length - 1, mid;
		 while (left <= right){
			 mid = (left + right) / 2;
			 if (A[mid] == target)
				 return  true;
			 else if (target < A[mid])
				 right = mid - 1;
			 else
				 left = mid + 1;
		 }
		 return false;
	 }
 };

“Here is the code that used binary search. The matrix can be viewed as a one-dimensional sorted array. The value of element i in this array in the matrix is matrix[i/cols][i%cols]“

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int rows = matrix.length;
        int cols = matrix[0].length;

        int s = 0, e = rows* cols - 1;
        while(s<=e){
            int m = s+(e-s)/2;
            if(matrix[m/cols][m%cols]==target)
            return true;
            else if(matrix[m/cols][m%cols]>target){
                e = m -1;
            }else{
                s = m + 1;
            }
        }
        return false;
    }
}

LeetCode之Search a 2D Matrix,布布扣,bubuko.com

LeetCode之Search a 2D Matrix

原文：http://blog.csdn.net/smileteo/article/details/22199087