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  • Search a 2D Matrix ——LeetCode

    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.

    题目大意:给一个m*n矩阵,每一行都是递增有序,逐行也是递增的,要求设计一个高效算法检查目标元素是否存在于此矩阵中。

    解题思路:可以把整个矩阵展开,就是一个长数组,数组长度为M*N,用二分查找即可,就是需要把二分查找的位置转化为矩阵下标。假设有row行,col列,那么key对应的矩阵中的元素应该是matrix[key/col][key%col],这样就转为二分查找了。

    Talk is cheap>>

       public boolean searchMatrix(int[][] matrix, int target) {
            if (matrix == null || matrix[0][0] > target)
                return false;
            int rowLen = matrix.length;
            int colLen = matrix[0].length;
            int low = 0, high = rowLen * colLen - 1;
    
            while (low <= high) {
                int mid = (low + high) >>> 1;
                int x = mid / colLen;
                int y = mid % colLen;
                if (matrix[x][y] > target) {
                    high = mid - 1;
                } else if (matrix[x][y] < target) {
                    low = mid + 1;
                } else {
                    return true;
                }
            }
            return false;
        }
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  • 原文地址:https://www.cnblogs.com/aboutblank/p/4394729.html
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