240. Search a 2D Matrix II

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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.

题目大意

在一个mxn的矩阵中，寻找目标，若存在，则返回true；若不存在，则返回false。
该矩阵有以下特点:

• 每一行从左到右递增。
• 每一列从上到下递增。

解题思路

利用该矩阵的特性，使用分治法的思想，从左下角开始找起，若该值小于目标值，则向右移动一个单位；若该值大于目标值，则向上移动一个单位。若超出边界，返回false。

算法复杂度

O(M+N)

代码实现

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