|
5 | 5 | ### 快速排序 |
6 | 6 |
|
7 | 7 | ```Python |
8 | | -def partition(A, start, end): |
9 | | - if start >= end: |
| 8 | +import random |
| 9 | + |
| 10 | +def partition(nums, left, right): |
| 11 | + if left >= right: |
10 | 12 | return |
11 | | - |
12 | | - l, r = start, end - 1 |
13 | | - while l < r: |
14 | | - while l < r and A[l] <= A[end]: |
15 | | - l += 1 |
16 | | - while l < r and A[r] >= A[end]: |
17 | | - r -= 1 |
18 | | - |
19 | | - A[l], A[r] = A[r], A[l] |
20 | | - |
21 | | - swap = r + int(A[r] < A[end]) |
22 | 13 |
|
23 | | - A[end], A[swap] = A[swap], A[end] |
| 14 | + pivot_idx = random.randint(left, right) |
| 15 | + pivot = nums[pivot_idx] |
| 16 | + |
| 17 | + nums[right], nums[pivot_idx] = nums[pivot_idx], nums[right] |
| 18 | + |
| 19 | + partition_idx = left |
| 20 | + for i in range(left, right): |
| 21 | + if nums[i] < pivot: |
| 22 | + nums[partition_idx], nums[i] = nums[i], nums[partition_idx] |
| 23 | + partition_idx += 1 |
| 24 | + |
| 25 | + nums[right], nums[partition_idx] = nums[partition_idx], nums[right] |
24 | 26 |
|
25 | | - partition(A, swap + 1, end) |
26 | | - partition(A, start, swap - 1) |
| 27 | + partition(nums, partition_idx + 1, right) |
| 28 | + partition(nums, left, partition_idx - 1) |
27 | 29 |
|
28 | 30 | return |
29 | 31 |
|
@@ -131,6 +133,67 @@ if __name__ == '__main__': |
131 | 133 | print(heapsort(a)) |
132 | 134 | ``` |
133 | 135 |
|
| 136 | +## 题目 |
| 137 | + |
| 138 | +### [kth-largest-element-in-an-array](https://leetcode-cn.com/problems/kth-largest-element-in-an-array/) |
| 139 | + |
| 140 | +思路 1: sort 后取第 k 个,最简单直接,复杂度 O(N log N) 代码略 |
| 141 | + |
| 142 | +思路 2: 使用最小堆,复杂度 O(N log k) |
| 143 | + |
| 144 | +```Python |
| 145 | +class Solution: |
| 146 | + def findKthLargest(self, nums: List[int], k: int) -> int: |
| 147 | + # note that in practice there is a more efficient python build-in function heapq.nlargest(k, nums) |
| 148 | + min_heap = [] |
| 149 | + |
| 150 | + for num in nums: |
| 151 | + if len(min_heap) < k: |
| 152 | + heapq.heappush(min_heap, num) |
| 153 | + else: |
| 154 | + if num > min_heap[0]: |
| 155 | + heapq.heappushpop(min_heap, num) |
| 156 | + |
| 157 | + return min_heap[0] |
| 158 | +``` |
| 159 | + |
| 160 | +思路 3: Quick select,方式类似于快排,每次 partition 后检查 pivot 是否为第 k 个元素,如果是则直接返回,如果比 k 大,则继续 partition 小于 pivot 的元素,如果比 k 小则继续 partition 大于 pivot 的元素。相较于快排,quick select 每次只需 partition 一侧,因此平均复杂度为 O(N) |
| 161 | + |
| 162 | +```Python |
| 163 | +class Solution: |
| 164 | + def findKthLargest(self, nums: List[int], k: int) -> int: |
| 165 | + |
| 166 | + k -= 1 # 0-based index |
| 167 | + |
| 168 | + def partition(left, right): |
| 169 | + pivot_idx = random.randint(left, right) |
| 170 | + pivot = nums[pivot_idx] |
| 171 | + |
| 172 | + nums[right], nums[pivot_idx] = nums[pivot_idx], nums[right] |
| 173 | + |
| 174 | + partition_idx = left |
| 175 | + for i in range(left, right): |
| 176 | + if nums[i] > pivot: |
| 177 | + nums[partition_idx], nums[i] = nums[i], nums[partition_idx] |
| 178 | + partition_idx += 1 |
| 179 | + |
| 180 | + nums[right], nums[partition_idx] = nums[partition_idx], nums[right] |
| 181 | + |
| 182 | + return partition_idx |
| 183 | + |
| 184 | + left, right = 0, len(nums) - 1 |
| 185 | + while True: |
| 186 | + partition_idx = partition(left, right) |
| 187 | + if partition_idx == k: |
| 188 | + return nums[k] |
| 189 | + elif partition_idx < k: |
| 190 | + left = partition_idx + 1 |
| 191 | + else: |
| 192 | + right = partition_idx - 1 |
| 193 | +``` |
| 194 | + |
| 195 | + |
| 196 | + |
134 | 197 | ## 参考 |
135 | 198 |
|
136 | 199 | [十大经典排序](https://www.cnblogs.com/onepixel/p/7674659.html) |
|
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