leetcode 074 Search a 2D Matrix

最新推荐文章于 2020-07-03 09:27:28 发布

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本文介绍了一种在具有特定排序属性的二维矩阵中查找目标值的高效算法。该算法利用了矩阵内元素的有序性质，通过分治策略快速定位目标值。

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.

/*class Solution {
public:
	bool searchMatrix(vector<vector<int>>& matrix, int target) {
		int rows = matrix.size(), cols = 0;

		if(rows == 0) return false;

		cols = matrix[0].size();

		if(target > matrix[rows-1][cols-1] || target < matrix[0][0]) return false;

		int l = 0, r = rows-1;

		while(l <= r) {
			int mid = (l+r)/2;
			if(matrix[mid][0] == target) return true;

			else if(matrix[mid][0] > target) {
				r=mid-1;
			} else {
				l = mid+1;
			}
		}

		int row=l-1;
		if(matrix[row][0] > target) row -= 1;
		l=0;
		r=cols-1;

		while(l <= r) {
			int mid = (l+r)/2;
			if(matrix[row][mid] == target) return true;
			else if(matrix[row][mid] > target) {
				r = mid-1;
			} else {
				l = mid+1;
			}
		}

		return false;
	}
}*/

//n*m matrix convert to an array, matrix[x][y] arr[x*m+y]
//an array convert to n*m matrix, a[x] matrix[x/m][x%m]
class Solution {
public:
	bool searchMatrix(vector<vector<int>> &matrix, int target) {
		int n = matrix.size();
		int m = matrix[0].size();

		int l=0,r=m*n-1;
		if(target>matrix[n-1][m-1] || target < matrix[0][0]) return false;
		while(l<=r) {
			int mid = (l+r)/2;
			if(matrix[mid/m][mid%m] == target) return true;
			else if(matrix[mid/m][mid%m] > target) {
				r=mid-1;
			} else {
				l=mid+1;
			}
		}

		return false;
	}
}