Divide and Conquer -- Leetcode problem240. Search a 2D Matrix II

• 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.
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.

• 分析：这道题是寻找一个数target在不在m*n数组中，如果在返回true，否则返回false。
• 思路一：直接使用for循环解答，无技术可言。

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {    int height = matrix.size();    if (height == 0) return false;    int width = matrix[0].size();    for (int i = 0; i < height; i++) {        for (int j = 0; j < width; j++) {            if (target == matrix[i][j])                return true;        }    }    return false;}};

• 思路二：由于数组中的元素是有一定排列顺序的，依据这一特性可以求解。

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {    int m_height = matrix.size();    if (m_height == 0) return false;    int m_width = matrix[0].size();    int width = 0;    int height = m_width - 1;    while (width < m_height && height >= 0) {        if (target == matrix[width][height]) return true;        else if (target < matrix[width][height]) height--;        else width++;    }    return false;}};