LeetCode 240. Search a 2D Matrix II

原题链接在这里：https://leetcode.com/problems/search-a-2d-matrix-ii/

题目：

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.

题解：

本题是每一行每一列都是升序，从右上角开始x = matrix[i][j], i = 0, j = matrix[0].length - 1, 用x 和target 比较，若是相等则返回true.

若是小于target, 说明肯定不会在这一行，所以i++. 若是大于target, 说明必然不会在这一列所以 j--.

若走到边界还没有返回说明没有这个target, 返回false.

Note: j的初始量是 j = n-1, 注意减一.

Time Complexity: O(m+n). m = matrix.length. n = matrix[0].length.

Space: O(1).

AC Java:

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return false;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        int i = 0;
        int j = n-1; //error
        while(i<m && j>=0){
            int x = matrix[i][j];
            if(x == target){
                return true;
            }else if(x < target){
                i++;
            }else{
                j--;
            }
        }
        return false;
    }
}

类似Search a 2D Matrix.

跟上Kth Smallest Element in a Sorted Matrix.