算法分析与设计第二周习题：分治算法之P215,P169,P53

题目链接
题目名称：215. Kth Largest Element in an Array
题目描述：
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

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python代码：

class Solution(object):    def findKthLargest(self, nums, k):        return recFind(nums, k)#选取数组中位于中间位置的值作为分水岭分界值#以分界值将数组分为小于，等于和大于该值的三部分，并统计三部分数组的长度#根据长度和k值来进行下一步的判定def recFind(nums, k):    length = len(nums)    target = nums[int(length / 2)]    if length == 1:        return nums[0]    smallerCount, biggerCount, equalCount = 0, 0, 0    smallerList, biggerList = [], []    for i in range(length):        if nums[i] < target:            smallerList.append(nums[i])            smallerCount += 1        elif nums[i] > target:            biggerList.append(nums[i])            biggerCount += 1        else:            equalCount += 1    #若大值数组长度大于等于k，则在大值数组中寻找kth最大值    if biggerCount >= k:        return recFind(biggerList, k)    #若kth最大值在等值数组中，则可直接返回分界值    elif biggerCount + equalCount >= k:        return target    #若Kth最大值在小值数组中，则减少k的值，并在小值数组中寻找    else:        return recFind(smallerList, k - biggerCount - equalCount)

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代码效率：

总结：
代码使用了老师课上所说的分治算法，并实现了较高的算法效率。
下面贴出了169题和53题的代码，这两道题难度为easy，就不再做详细分析。

169题

int recMaj(vector<int>& nums, int left, int right) {    if (left == right)        return nums[left];    int mid = (left + right) / 2;    int maj1 = recMaj(nums, left, mid);    int maj2 = recMaj(nums, mid + 1, right);    if (maj1 == maj2)        return maj1;    else {        int count1 = 0, count2 = 0;        for (int i = left; i <= right; i++) {            if (nums[i] == maj1)                count1++;            else if (nums[i] == maj2)                count2++;        }        return count1 < count2 ? maj2 : maj1;    }}class Solution {public:    int majorityElement(vector<int>& nums) {        return recMaj(nums, 0, nums.size() - 1);    }};

53题

class Solution(object):    def maxSubArray(self, nums):        return divide(nums)def divide(nums):    if len(nums) == 1:        return nums[0]    length = len(nums)    mid = int(length / 2)    #将数组划分为两半    #并得到两半分别的最大值    max1 = divide(nums[0:mid])    max2 = divide(nums[mid:length])    #计算含边界时的最大值    maxLBorder,sumLBorder, maxRBorder, sumRBorder = 0, 0, 0, 0    for i in range(mid, -1, -1):        sumLBorder += nums[i]        if sumLBorder > maxLBorder:            maxLBorder = sumLBorder    for i in range(mid + 1, length):        sumRBorder += nums[i]        if sumRBorder > maxRBorder:            maxRBorder = sumRBorder    #若数组全为负数，则直接返回两半中的最大值    #不考虑边界，是因为把边界最小值设置为0,    #当数组全为负数时，若考虑边界则程序会返回0，这是错误的    if (max1 < 0 and max2 < 0):        return max(max1, max2)    return max(max1, max2, maxLBorder + maxRBorder)