[leetcode-查找]--74. Search a 2D Matrix

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Question 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:

(1) Integers in each row are sorted from left to right.
(2)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.

中文：对于一个整型二维数组，每一行都是升序有序的，下一行第一个数比上一行最后一个数大。对于一个给定的目标数target 判断是否在二维数组中，若在返回true，不在返回false。

实现：这里我首先遍历每一行的第一个数据，获取到target应该在哪一行进行检验。然后在对应的行使用二分查找，找到数据。对于m*n 的数组，时间复杂度是O(m*logn)。

实现源码：

/** * 先找列, 遍历. 找行用二分查找 * @param matrix * @param target * @return */public static boolean searchMatrix(int[][] matrix, int target) {    if(matrix == null || matrix.length==0 || matrix[0].length==0){        return false;    }    int rows = matrix.length;//总行数    int columns = matrix[0].length;//总列数    int row=0;//比较的行数    int i=0;    //找到比较行    for(i=0; i<rows; i++){        if( (matrix[i][0] <= target) && (i < rows-1) && (matrix[i+1][0] > target)){            row = i;            break;        }        if( (matrix[i][0] <= target) && ( i == rows-1)){            row = i;            break;        }    }    //二分查找找到对应的位置,找不到返回false    int lo=0, hi=matrix[row].length-1;    int mid;//中间索引    while (lo <= hi){        mid = (lo+hi)/2;        if(matrix[row][mid] == target){            return true;        } else if(matrix[row][mid] < target){            lo = mid+1;        }else{            hi = mid-1;        }    }    return false;}

本文完整代码 github地址：
https://github.com/leetcode-hust/leetcode/blob/master/louyuting/src/leetcode/Question74.java

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