[Leetcode] Search a 2D Matrix 搜索二维矩阵

Search a 2D Matrix I

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.

二分法

复杂度

时间 O(log(MN)) 空间 O(1)

思路

我们可以把二维数组想象成一个一维数组，第一行尾连着第二行头，第二行尾连着第三行头...同样是有个最小值最大值，二分得到中间值，通过对中间值取模可以得到对应二维数组的下标。这样，该题就转化为了一维有序数组二分搜索的题了。

注意

• 二分搜索的几个边界条件是min<=max，min=mid+1,max=mid-1

代码

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        if(m == 0) return false;
        int n = matrix[0].length;
        int min = 0, max = m * n - 1;
        while(min <= max){
            int mid = min + (max - min) / 2;
            int row = mid / n;
            int col = mid % n;
            if(matrix[row][col] == target){
                return true;
            } else if (matrix[row][col] < target){
                min = mid + 1;
            } else {
                max = mid - 1;
            }
        }
        return false;
    }
}

2018/2

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:
            return False
        rows = len(matrix)
        cols = len(matrix[0])
        min, max = 0, rows * cols - 1
        while min <= max:
            mid = min + (max - min) // 2
            row = mid // cols
            col = mid % cols
            if matrix[row][col] == target:
                return True
            elif matrix[row][col] < target:
                min = mid + 1
            elif matrix[row][col] > target:
                max = mid - 1
        return False

Search a 2D Matrix II

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 in ascending from left to right. Integers in each column are sorted in ascending from top to bottom. For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

贪心法

复杂度

时间 O(N+M) 空间 O(1)

思路

虽说该方法叫贪心法不太得当，但是贪心最能描述该方法的过程。由于数组的特性，我们可以从二维数组的右上角出发，如果目标小于该数，则向左移动（左边的数字肯定更小，而当前列中所有数字都比该数字大）。如果该数比目标小，则向下移动（下边的数字肯定更大，而当前行的所有数字逗比该数字小）。这样反复重复该过程就能找到目标数。如果直到左下角都没有该数，说明找不到该数。同样的，这题也可以从左下角向右上角寻找。

代码

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0){
            return false;
        }
        int i = 0, j = matrix[0].length - 1;
        while(i < matrix.length && j >= 0){
            int curr = matrix[i][j];
            if(curr == target){
                return true;
            // 比目标小则向下
            } else if(curr > target){
                j--;
            // 比目标大则向左
            } else {
                i++;
            }
        }
        return false;
    }
}

2018/2

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:
            return False
        rows = len(matrix)
        cols = len(matrix[0])
        row, col = 0, cols - 1
        while row >= 0 and row < rows and col >=0 and col < cols:
            if matrix[row][col] > target:
                col -= 1
            elif matrix[row][col] < target:
                row += 1
            elif matrix[row][col] == target:
                return True
        return False