原題
給定一組 n 個(gè)不同大小的 nuts 和 n 個(gè)不同大小的 bolts。nuts 和 bolts 一一匹配。 不允許將 nut 之間互相比較，也不允許將 bolt 之間互相比較。也就是說(shuō)，只許將 nut 與 bolt 進(jìn)行比較， 或?qū)?bolt 與 nut 進(jìn)行比較。請(qǐng)比較 nut 與 bolt 的大小。
樣例
Nuts 用一個(gè)string數(shù)組表示 nuts [] = ['ab','bc','dd','gg']. Bolts 也用一個(gè)string數(shù)組表示 bolts[] = ['AB','GG', 'DD', 'BC']. 我們將提供一個(gè)比較函數(shù)，以比較 nut 與 bolt 的大小。將 nuts 進(jìn)行升序排序，使得 nuts 與 bolts 位置對(duì)等。
比如 nuts = ['ab','bc','dd','gg'], bolts = ['BC','AA','DD','GG']
解題思路
- Selection Sort - 利用partition的思想
因?yàn)镹uts之間，Bolts之間都是不能相互比較的，所以思路就是隨機(jī)選擇一個(gè)Nut來(lái)partition Bolts然后再反過(guò)來(lái)用那個(gè)已經(jīng)在正確位置的Bolt來(lái)partition Nuts - 初始：cc aa dd bb
- 隨機(jī)找到BB，partition之后
- aa bb cc dd （cc, dd的順序不確定）
- 之后再拿bb去反過(guò)來(lái)partition，得到
- AA BB DD CC
- 遞歸向下，分別左右sort
- 注意比較的時(shí)候compare函數(shù)只能
- partition函數(shù)的執(zhí)行過(guò)程，模擬
pivot是4，通過(guò)for循環(huán)找到，交換到第一位，并記住pivot_partner = 4
開(kāi)始：[4, 3, 2, 9, 5]
5 > 4: [5, 3, 2, 9, 5]
3 < 4: [5, 3, 2, 9, 3]
9 > 4: [5, 9, 2, 9 , 3]
2 < 4: [5, 9, 2, 2, 3]
此時(shí)L在二號(hào)位，讓nums[L] = pivot_partner
[5, 9, 4, 2, 3] =>保證了4前面都是大于4的，后面都是小于4的
完整代碼
# class Comparator:
# def cmp(self, a, b)
# You can use Compare.cmp(a, b) to compare nuts "a" and bolts "b",
# if "a" is bigger than "b", it will return 1, else if they are equal,
# it will return 0, else if "a" is smaller than "b", it will return -1.
# When "a" is not a nut or "b" is not a bolt, it will return 2, which is not valid.
class Solution:
# @param nuts: a list of integers
# @param bolts: a list of integers
# @param compare: a instance of Comparator
# @return: nothing
def sortNutsAndBolts(self, nuts, bolts, compare):
# write your code here
if nuts == [] or bolts == [] or compare is None:
return
self.compare = compare
self.quickSort(nuts, bolts, 0, len(nuts) - 1)
def quickSort(self, nuts, bolts, left, right):
if left >= right:
return
split_nut_pos = self.partition(nuts, bolts[left], left, right)
split_bolt_pos = self.partition(bolts, nuts[split_nut_pos], left, right)
self.quickSort(nuts, bolts, left, split_nut_pos - 1)
self.quickSort(nuts, bolts, split_nut_pos + 1, right)
def partition(self, items, pivot, left, right):
if items == [] or pivot == None:
return
# 首先把pivot換到數(shù)組的第一個(gè)位置
for i in range(left, right + 1):
if self.compare.cmp(pivot, items[i]) == 0 \
or self.compare.cmp(items[i], pivot) == 0:
items[left], items[i] = items[i], items[left]
break
# 如果pivot是AA,此時(shí)pivot_partner就是aa
pivot_partner = items[left]
while left < right:
while left < right and \
(self.compare.cmp(pivot, items[right]) == -1 \
or self.compare.cmp(items[right], pivot) == 1):
right -= 1
items[left] = items[right]
while left < right and \
(self.compare.cmp(pivot, items[left]) == 1 \
or self.compare.cmp(items[left], pivot) == -1):
left += 1
items[right] = items[left]
items[left] = pivot_partner
return left
