leetcode--Search a 2D Matrix

最新推荐文章于 2019-04-06 12:59:12 发布

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最新推荐文章于 2019-04-06 12:59:12 发布

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分类专栏：算法

本文链接：https://blog.youkuaiyun.com/u014491519/article/details/78272222

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本文介绍了一种高效的搜索算法，用于在一特殊性质的二维矩阵中查找特定值。该矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了两种解决方案：一种通过定位目标值所在的行再进行折半查找；另一种则是将二维矩阵视为一维数组进行折半查找。

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题目

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.

解法1：有序数组的折半查找
先找出target属于哪一行，然后在此行中进行查找

public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0 || matrix[0].length == 0)
            return false;
        int start = 0, end = matrix.length-1;
        if(target < matrix[start][0] || target > matrix[matrix.length-1][matrix[0].length-1])
            return false;
        if(target >= matrix[end][0])
            start = end;
        else{
            while(end - start > 1){
                int mid = (end + start)/2;
                if(target >= matrix[mid][0] && target < matrix[end][0])
                    start = mid;
                else
                    end = mid;
            }
        }
        int row = start;
        start = 0;
        end = matrix[0].length-1;
        while(start <= end){
            int mid = (start+end)/2;
            if(target < matrix[row][mid])
                end = mid - 1;
            else if(target > matrix[row][mid])
                start = mid + 1;
            else
                return true;
        }
        return false;
    }

解法2
将二维矩阵看作是一维数组，进行折半查找
对应关系：

m*n matrix[i][j] == array[i*n + j]
array[x] == matrix[x/n][x%n]
折半查找

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