240.search-a-2-d-matrix-ii.md

题目地址

https://leetcode.com/problems/search-a-2d-matrix-ii/description/

题目描述

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.
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.

思路

符合直觉的做法是两层循环遍历，时间复杂度是O(m * n), 有没有时间复杂度更好的做法呢？ 答案是有，那就是充分运用矩阵的特性（横向纵向都递增）， 我们可以从角落（左下或者右上）开始遍历，这样时间复杂度是O(m + n).

其中蓝色代表我们选择的起点元素， 红色代表目标元素。

关键点解析

• 从角落开始遍历，利用递增的特性简化时间复杂度

代码

代码支持：JavaScript, Python3

JavaScript Code:

/*
 * @lc app=leetcode id=240 lang=javascript
 *
 * [240] Search a 2D Matrix II
 *
 * https://leetcode.com/problems/search-a-2d-matrix-ii/description/
 *
 * 
 */
/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function(matrix, target) {
    if (!matrix || matrix.length === 0) return 0;

    let colIndex = 0;
    let rowIndex = matrix.length - 1;
    while(rowIndex > 0 && target < matrix[rowIndex][colIndex]) {
        rowIndex --;
    }

    while(colIndex < matrix[0].length) {
        if (target === matrix[rowIndex][colIndex]) return true;
        if (target > matrix[rowIndex][colIndex]) {
            colIndex ++;
        } else if (rowIndex > 0){
            rowIndex --;
        } else {
            return false;
        }
    }

    return  false;
};

Python Code:

class Solution:
    def searchMatrix(self, matrix, target):
        m = len(matrix)
        if m == 0:
            return False
        n = len(matrix[0])
        i = m - 1
        j = 0

        while i >= 0 and j < n:
            if matrix[i][j] == target:
                return True
            if matrix[i][j] > target:
                i -= 1
            else:
                j += 1
        return False