2019-11-02 12:07:41 +08:00
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# 4. 二维数组中的查找
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2019-11-02 14:18:29 +08:00
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## 题目链接
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[牛客网](https://www.nowcoder.com/practice/abc3fe2ce8e146608e868a70efebf62e?tpId=13&tqId=11154&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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2019-11-02 12:07:41 +08:00
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## 题目描述
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给定一个二维数组，其每一行从左到右递增排序，从上到下也是递增排序。给定一个数，判断这个数是否在该二维数组中。
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```html
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Consider the following matrix:
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[
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[1, 4, 7, 11, 15],
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[2, 5, 8, 12, 19],
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[3, 6, 9, 16, 22],
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[10, 13, 14, 17, 24],
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[18, 21, 23, 26, 30]
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]
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Given target = 5, return true.
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Given target = 20, return false.
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```
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## 解题思路
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要求时间复杂度 O(M + N)，空间复杂度 O(1)。其中 M 为行数，N 为 列数。
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2020-11-04 01:20:57 +08:00
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该二维数组中的一个数，小于它的数一定在其左边，大于它的数一定在其下边。因此，从右上角开始查找，就可以根据 target 和当前元素的大小关系来快速地缩小查找区间，每次减少一行或者一列的元素。当前元素的查找区间为左下角的所有元素。
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2019-11-02 12:07:41 +08:00
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2019-12-06 10:11:23 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/35a8c711-0dc0-4613-95f3-be96c6c6e104.gif" width="400px"> </div><br>
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2019-11-02 12:07:41 +08:00
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```java
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public boolean Find(int target, int[][] matrix) {
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
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return false;
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int rows = matrix.length, cols = matrix[0].length;
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int r = 0, c = cols - 1; // 从右上角开始
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while (r <= rows - 1 && c >= 0) {
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if (target == matrix[r][c])
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return true;
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else if (target > matrix[r][c])
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r++;
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else
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c--;
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}
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return false;
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}
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```
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