#leetcode #search a 2D Matrix II

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted in ascending from left to right.
• Integers in each column is 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 , [+], [approx., +, +]] 

given target  = 5 , Return true .

Given Target = 20 , return false .

Analysis: This question just saw on EPI the day before yesterday ....

More than the search a 2D matrix, that is, each column is monotonically increasing, assuming that the Matrix is n * N, from the upper right corner of the element to start judging, set to X,

if x = = target, returns True

If x < target, go to the left column to find

If x > Target, you want to find the following line

Time complexity O (M + N)

public class Solution {public    Boolean Searchmatrix (int[][] matrix, int target) {        if (Matrix = = NULL | | matrix.leng th = = 0 | | Matrix[0].length = = 0) {            return false;        }                    int r = 0;        int c = matrix[0].length-1;        while (R < matrix.length && C >= 0) {            if (matrix[r][c] = = target) {                return true;            } else if (Matrix[r][c] > target) {                c--;            } else{                r++;            }        }                return false;    }}

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#leetcode #search a 2D Matrix II