|
| 1 | +## [0074. 搜索二维矩阵](https://leetcode.cn/problems/search-a-2d-matrix/) |
| 2 | + |
| 3 | +- 标签:数组、二分查找、矩阵 |
| 4 | +- 难度:中等 |
| 5 | + |
| 6 | +## 题目大意 |
| 7 | + |
| 8 | +**描述**:给定一个 `m * n` 大小的有序二维矩阵 `matrix`。矩阵中每行元素从左到右升序排列,每列元素从上到下升序排列。再给定一个目标值 `target`。 |
| 9 | + |
| 10 | +**要求**:判断矩阵中是否存在目标值 `target` |
| 11 | + |
| 12 | +**说明**: |
| 13 | + |
| 14 | +- $m == matrix.length$。 |
| 15 | +- $n == matrix[i].length$。 |
| 16 | +- 1ドル \le m, n \le 100$。 |
| 17 | +- $-10^4 \le matrix[i][j], target \le 10^4$。 |
| 18 | + |
| 19 | +**示例**: |
| 20 | + |
| 21 | +```Python |
| 22 | +输入 matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3 |
| 23 | +输出 True |
| 24 | +``` |
| 25 | + |
| 26 | +## 解题思路 |
| 27 | + |
| 28 | +### 思路 1:二分查找 |
| 29 | + |
| 30 | +二维矩阵是有序的,可以考虑使用二分搜索来进行查找。 |
| 31 | + |
| 32 | +1. 首先二分查找遍历对角线元素,假设对角线元素的坐标为 `(row, col)`。把数组元素按对角线分为右上角部分和左下角部分。 |
| 33 | +2. 然后对于当前对角线元素右侧第 `row` 行、对角线元素下侧第 `col` 列进行二分查找。 |
| 34 | + 1. 如果找到目标,直接返回 `True`。 |
| 35 | + 2. 如果找不到目标,则缩小范围,继续查找。 |
| 36 | + 3. 直到所有对角线元素都遍历完,依旧没找到,则返回 `False`。 |
| 37 | + |
| 38 | +### 思路 1:代码 |
| 39 | + |
| 40 | +```Python |
| 41 | +class Solution: |
| 42 | + # 二分查找对角线元素 |
| 43 | + def diagonalBinarySearch(self, matrix, diagonal, target): |
| 44 | + left = 0 |
| 45 | + right = diagonal |
| 46 | + while left < right: |
| 47 | + mid = left + (right - left) // 2 |
| 48 | + if matrix[mid][mid] < target: |
| 49 | + left = mid + 1 |
| 50 | + else: |
| 51 | + right = mid |
| 52 | + return left |
| 53 | + |
| 54 | + def rowBinarySearch(self, matrix, begin, cols, target): |
| 55 | + left = begin |
| 56 | + right = cols |
| 57 | + while left < right: |
| 58 | + mid = left + (right - left) // 2 |
| 59 | + if matrix[begin][mid] < target: |
| 60 | + left = mid + 1 |
| 61 | + elif matrix[begin][mid] > target: |
| 62 | + right = mid - 1 |
| 63 | + else: |
| 64 | + left = mid |
| 65 | + break |
| 66 | + return begin <= left <= cols and matrix[begin][left] == target |
| 67 | + |
| 68 | + def colBinarySearch(self, matrix, begin, rows, target): |
| 69 | + left = begin + 1 |
| 70 | + right = rows |
| 71 | + while left < right: |
| 72 | + mid = left + (right - left) // 2 |
| 73 | + if matrix[mid][begin] < target: |
| 74 | + left = mid + 1 |
| 75 | + elif matrix[mid][begin] > target: |
| 76 | + right = mid - 1 |
| 77 | + else: |
| 78 | + left = mid |
| 79 | + break |
| 80 | + return begin <= left <= rows and matrix[left][begin] == target |
| 81 | + |
| 82 | + def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: |
| 83 | + rows = len(matrix) |
| 84 | + if rows == 0: |
| 85 | + return False |
| 86 | + cols = len(matrix[0]) |
| 87 | + if cols == 0: |
| 88 | + return False |
| 89 | + |
| 90 | + min_val = min(rows, cols) |
| 91 | + index = self.diagonalBinarySearch(matrix, min_val - 1, target) |
| 92 | + if matrix[index][index] == target: |
| 93 | + return True |
| 94 | + for i in range(index + 1): |
| 95 | + row_search = self.rowBinarySearch(matrix, i, cols - 1, target) |
| 96 | + col_search = self.colBinarySearch(matrix, i, rows - 1, target) |
| 97 | + if row_search or col_search: |
| 98 | + return True |
| 99 | + return False |
| 100 | +``` |
0 commit comments