LeetCode--No.74--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.

我的代码：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0 || matrix == null)
            return false;
        int m = matrix.length;
        int n = matrix[0].length;
        if (m < 1 || n < 1)
            return false;
        int start = 0;
        int end = m - 1;
        int row_index = 0;
        while(start <= end){
            int mid = (start + end) / 2;
            if (target < matrix[mid][0])
                end = mid - 1;
            else if (target > matrix[mid][n - 1])
                start = mid + 1;
            else{
                row_index = mid;
                break;
            }
        }
        start = 0;
        end = n - 1;
        while(start <= end){
            int mid = (start + end) / 2;
            if (target == matrix[row_index][mid])
                return true;
            else if (target < matrix[row_index][mid])
                end = mid - 1;
            else
                start = mid + 1;
        }
        return false;
    }
}

就是两个，二分查找。时间复杂度没问题。都是O(logn). 然而看了下别人的代码才发现自己的写的有多不简洁。
就是感觉思维还是很定式。觉得按顺序的查找，肯定是二分。那二维的自然就是两次二分。然而其实不需要。大可以把二维数组当做一维。
下面是优化后的代码-- 抄自 leetcode solution part

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0)
            return false;
        int start = 0, rows = matrix.length, cols = matrix[0].length; 
        int end = rows*cols - 1;
        while(start <= end){
            int mid = (start + end) / 2;
            if (matrix[mid/cols][mid%cols] == target)
                return true;
            if (matrix[mid/cols][mid%cols] < target)
                start = mid + 1;
            else
                end = mid - 1;
        }
        return false;
    }
}

我今天真的撑死了。

我要坚持住早起去健身房锻炼。我要成为健康的宝宝。撑死我了。

再刷两道题消化消化，再去吃冻酸奶！

LeetCode--No.74--Search a 2D Matrix

猜你喜欢