【LeetCode-面試演算法經典-Java實現】【074-Search a 2D Matrix(搜尋二維矩陣)】,leetcode--java
【074-Search a 2D Matrix(搜尋二維矩陣)】【LeetCode-面試演算法經典-Java實現】【所有題目目錄索引】原題
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: Given target = 3, return true.
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
題目大意
給定一個二維矩陣,實現一個演算法在矩陣中實現快速搜尋。即給定k,在矩陣中搜尋k
矩陣中下面的性質:每一行每一列都是排好序的,每一行的第一個數都比上一行的最後一個數大。
解題思路
解法一:先用二叉查看找演算法找到數字所在的列,再用二叉尋找演算法找數字所在的列。找到就返回true,否則返回false。解法二:見【劍指Offer學習】【面試題3 :二維數組中的尋找】
代碼實現
演算法實作類別
public class Solution { public boolean searchMatrix(int[][] matrix, int target) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { return false; } int row = matrix.length; int column = matrix[0].length; int low = 0; int high = row - 1; int mid = 0; // 找結果所在的列 while (low <= high) { mid = low + (high - low) / 2; if (target < matrix[mid][column - 1]) { high = mid - 1; } else if (target > matrix[mid][column - 1]) { low = mid + 1; } else { return true; } } // 決定列所在的最終位置 int targetRow = mid; if (matrix[mid][column - 1] < target) { targetRow++; } // 目標列超出,無結果 if (targetRow >= row) { return false; } low = 0; high = column - 1; // 找所在的行,找到返回true,沒有返回false while (low <= high) { mid = low + (high - low) / 2; if (target < matrix[targetRow][mid]) { high = mid - 1; } else if (target > matrix[targetRow][mid]) { low = mid + 1; } else { return true; } } return false; }}
評測結果
點擊圖片,滑鼠不釋放,拖動一段位置,釋放後在新的視窗中查看完整圖片。
特別說明
歡迎轉載,轉載請註明出處【http://blog.csdn.net/derrantcm/article/details/47142931】
著作權聲明:本文為博主原創文章,未經博主允許不得轉載。