# 240. LeetCode 240. Search a 2D Matrix II
[LeetCode 240. Search a 2D Matrix II](https://leetcode.com/problems/search-a-2d-matrix-ii/) (<font color=#FFB800>Medium</font> 53.3%)

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(<font color=#00AF9B>Easy</font> 53.8%)
(<font color=#FFB800>Medium</font> 39.6%)
(<font color=#FF375F>Hard</font>)
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- 限制 :

<ul>
<li><code>m == matrix.length</code></li>
<li><code>n == matrix[i].length</code></li>
<li><code>1 <= n, m <= 300</code></li>
<li><code>-10^9 <= matrix[i][j] <= 10^9</code></li>
<li><code>All the integers in each row are sorted in ascending order.</code></li>
<li><code>All the integers in each column are sorted in ascending order.</code></li>
<li><code>-10^9 <= target <= 10^9</code></li>
</ul>

- Solution
- 時間複雜度: $O(M+N)$
- 空間複雜度: $O(1)$

- 程式碼

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

</details>