LeetCode | Search a 2D Matrix

题目

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.

原来做过：九度Online Judge | 剑指offer | 题目1384：二维数组中的查找，文中给出了大牛的链接。

解法1：二分法

解法2：从右上或左下开始搜索，以右上为例，如果被搜索的元素大于target就减横坐标，如果小于target就减纵坐标，否则就是相等返回true

解法1

public class SearchA2DMatrix {
	public boolean searchMatrix(int[][] matrix, int target) {
		if (matrix == null || matrix.length == 0) {
			return false;
		}
		int M = matrix.length;
		int N = matrix[0].length;
		int low = 0;
		int high = M * N - 1;
		while (low <= high) {
			int mid = low + (high - low) / 2;
			if (matrix[mid / N][mid % N] == target) {
				return true;
			} else if (matrix[mid / N][mid % N] > target) {
				high = mid - 1;
			} else {
				low = mid + 1;
			}
		}
		return false;
	}
}

public class SearchA2DMatrix {
	public boolean searchMatrix(int[][] matrix, int target) {
		if (matrix == null || matrix.length == 0) {
			return false;
		}
		int M = matrix.length;
		int N = matrix[0].length;
		int i = 0;
		int j = N - 1;
		while (i <= M - 1 && j >= 0) {
			if (matrix[i][j] == target) {
				return true;
			} else if (matrix[i][j] > target) {
				--j;
			} else {
				++i;
			}
		}
		return false;
	}
}

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LeetCode | Search a 2D Matrix

原文：http://blog.csdn.net/perfect8886/article/details/21251873