数组_leetcode215

#coding=utf-8
# 解题思路：快速排序  20190302 找工作期间
# 关键点 index = partion(xx) ,index表示的是在整个数组中的位置

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """

        length = len(nums)

        if not nums or k <= 0 or k > length:
            return

        start = 0
        end  = length-1
        self.fastSort(nums,start,end)

        return nums[length-k]

    def fastSort(self,alist,start,end):
        if end <= start:
            return

        base = alist[start]
        index1 , index2 = start,end

        while start < end :

            while start < end and alist[end] >= base:
                end -= 1
            alist[start] = alist[end]

            while start < end and alist[start] <= base:
                start += 1
            alist[end] = alist[start]

        alist[start] = base

        self.fastSort(alist,index1,start-1)
        self.fastSort(alist,start+1,index2)



class Solution2(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """

        length = len(nums)

        if not nums or k <= 0 or k > length:
            return

        start = 0
        end  = length-1
        target = length-k

        index = self.partition(nums,start,end)

        while index != target:
            if index > target:
                index = self.partition(nums,start,index-1)
            else:
                index = self.partition(nums,index+1,end)

        return nums[target]

    def partition(self,alist,start,end):

        if end <= start:
            return start

        base = alist[start]
        index1 , index2 = start , end

        while start < end :
            while start < end and alist[end] >= base :
                end -= 1
            alist[start] = alist[end]

            while start < end  and alist[start] <= base:
                start += 1
            alist[end] = alist[start]

        alist[start] = base
        return start



nums1 = [3,2,1,5,6,4]
k1 = 2

nums2 = [3,2,3,1,2,4,5,5,6]
k2 = 4


s = Solution2()

print s.findKthLargest(nums2,k2)