[leetcode] quick sort (quick sort) (python)

Please see the explanation: quick sort - super detailed explanation of python version

### 所有的交换都是跟哨兵换的。
def quick_sort(nums, l, r):
    if l >= r: # >= 
        return 
    i = partition(nums, l, r) # 返回值是哨兵元素下标
    quick_sort(nums, l, i - 1)
    quick_sort(nums, i + 1, r)
    
def partition(nums, l, r):
    base = nums[l] # 哨兵
    i, j = l, r
    while i < j:
        while i < j and nums[j] > base: j -= 1
        nums[i] = nums[j]
        while i < j and nums[i] < base: i += 1
        nums[j] = nums[i]
    
    nums[i] = base # 此时 i == j, 将哨兵放到此位置
    
    return i

# nums = [5,7,1,8,3]
nums = [1,2] # quick_sort(nums, 0, -1)
quick_sort(nums, 0, len(nums)-1)
print(nums)

### 只有最后一个确定的位置的时候，才是跟哨兵换的。大部分都是nums[i], nums[j] 交换的
def quick_sort(nums, l, r):
    # 子数组长度为 1 时终止递归
    if l >= r: return
    # 哨兵划分操作
    i = partition(nums, l, r)
    # 递归左（右）子数组执行哨兵划分
    quick_sort(nums, l, i - 1)
    quick_sort(nums, i + 1, r)
    
def partition(nums, l, r):
    # 以 nums[l] 作为基准数 ############################################
    i, j = l, r
    while i < j:
        while i < j and nums[j] >= nums[l]: j -= 1
        while i < j and nums[i] <= nums[l]: i += 1
        nums[i], nums[j] = nums[j], nums[i]
    nums[l], nums[i] = nums[i], nums[l]
    return i

# 调用
nums = [3, 4, 1, 5, 2]
quick_sort(nums, 0, len(nums) - 1)
print(nums)