74. Search a 2D Matrix.md

74. Search a 2D Matrix

Question

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.

Example
Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Thinking:

• Method1:从右上角开始遍历
  • 如果temp大于target，左移。
  • 如果temp小于target，下移。
  • 如果等于，返回。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0) return false;
        int height = matrix.length;
        int width = matrix[0].length;
        int row = 0, col = width - 1;
        while(row < height && col >= 0){
            int temp = matrix[row][col];
            if(temp == target) return true;
            else if(temp < target)
                row++;
            else
                col--;
        }
        return false;
    }
}

二刷

这道题还是比较经典，二刷的完全可以记住一刷的结果。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int height = matrix.length, width = matrix[0].length;
        int col = width - 1, row = 0;
        while(row < height && col >= 0){
            if(matrix[row][col] == target) return true;
            else if(matrix[row][col] > target) col--;
            else row++;
        }
        return false;
    }
}

Third Time

• Method 1: Binary Search, AC 100%

   class Solution {
   	public boolean searchMatrix(int[][] matrix, int target) {
   		if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
   		int height = matrix.length, width = matrix[0].length;
   		int row = 0;
   		while(row < height && target > matrix[row][width - 1]){
   			row++;
   		}
   		return row >= height ? false: bs(matrix[row], 0, width - 1, target);
   	}
   	private boolean bs(int[] nums, int low, int high, int target){
   		if(low > high) return false;
   		int mid = low + (high - low) / 2;
   		if(target == nums[mid]) return true;
   		else if(target > nums[mid]) return bs(nums, mid + 1, high, target);
   		else return bs(nums, low, mid - 1, target);
   	}
   }

Amazon session

• Method 1: search

   class Solution {
   	public boolean searchMatrix(int[][] matrix, int target) {
   		if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
   		int height = matrix.length, width = matrix[0].length;
   		int x = 0, y = width - 1;
   		while(x >= 0 && x < height && y >= 0 && y < width){
   			if(matrix[x][y] == target) return true;
   			else if(matrix[x][y] < target){
   				x++;
   			}else{
   				y--;
   			}
   		}
   		return false;
   	}
   }