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  • Leetcode 240. Search a 2D Matrix II

    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 in ascending from left to right.
    • Integers in each column are sorted in ascending from top to bottom.

    For example,

    Consider the following matrix:

    [
      [1,   4,  7, 11, 15],
      [2,   5,  8, 12, 19],
      [3,   6,  9, 16, 22],
      [10, 13, 14, 17, 24],
      [18, 21, 23, 26, 30]
    ]
    

    Given target = 5, return true.

    Given target = 20, return false.


    解题思路:

    从左下角往右上角找,若是小于target就往右找,若是大于target就往上找。时间复杂度O(m+n)  n 为行数,m为列数。

    例如找136

    但Lintcode上类似题目问题变成找出多少个target,循环中稍微变化下,设置一个count, 就可以了。


    Java code:

    public class Solution {
        public boolean searchMatrix(int[][] matrix, int target) {
            //check corner case
            if(matrix == null || matrix.length == 0) {
                return false;
            }
            if(matrix[0] == null || matrix[0].length == 0) {
                return false;
            }
            //find from bottom left to top right
            int n = matrix.length; //row
            int m = matrix[0].length; //column
            int x = n-1;
            int y = 0;while ( x >= 0 && y < m) {
                if(matrix[x][y] < target) {
                    y++;
                } else if (matrix[x][y] > target) {
                    x--;
                } else {
                    return true;
                }
            } 
            return false;
        }
    }

    Reference:

    1. http://www.jiuzhang.com/solutions/search-a-2d-matrix-ii/

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  • 原文地址:https://www.cnblogs.com/anne-vista/p/5135390.html
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