leetcode--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.

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
       boolean exists = false;
		if(matrix.length > 0){
			int top = 0, bottom = matrix.length - 1;
			while(top < bottom){
				if(matrix[top + (bottom - top) / 2][0] > target)
					bottom = top + (bottom - top) / 2 - 1;
				else if(matrix[top + (bottom - top) / 2][0] == target){
					exists = true;
					break;
				}
				else{
					if(matrix[top + (bottom - top) / 2][matrix[0].length - 1] < target)
						top = top + (bottom - top) / 2 + 1;
					else{
						top = top + (bottom - top) / 2;
						break;
					}
				}
			}
			if(!exists)
				exists = (Arrays.binarySearch(matrix[top], target) < 0) ? false : true;
		}
		return exists;  
    }
}

  

原文地址:https://www.cnblogs.com/averillzheng/p/3802307.html