Leetcode 215. Kth Largest Element in an Array(第二周作业)
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先贴原题:
题意为:给出一个无序数组,然后指定一个整数值K,我们需要找出第k大的元素。
使用书上给出的方法:随便选择一个元素(我选择第size/2的元素),然后遍历数组,比这个元素小的丢到左边,比这个元素大的丢到右边,与元素相等的也记录下来。形成三个数组,倘若k小于等于右边数组的元素个数,则直接接着右边数组里找即可。倘若k比右边数组个数大,且小于等于右边数组的元素个数加中间数组元素的个数,则直接返回该元素的值。倘若大于右边数组和中间数组的元素个数的和,则在左边数组里找
class Solution {public:void findNum(vector<int> nums, int numsSize, int select) {left.clear();equal.clear();right.clear();for (int i = 0; i < numsSize; i++) {if (nums[i] < select) left.push_back(nums[i]);else if (nums[i] == select) equal.push_back(nums[i]);else right.push_back(nums[i]);}}int findKthLargest(vector<int> nums, int k) { //使用中间的元素作为标准int select = nums[nums.size()>>1];int result = 0;findNum(nums, nums.size(), select); //确定下一步在哪个数组里找if (k <= right.size()) result = findKthLargest(right, k);else if (k > right.size() && k <= equal.size() + right.size()) result = select;else result = findKthLargest(left, k - equal.size() - right.size());return result;}private:vector<int> left, equal, right;};
该算法的时间复杂度为O(nlogn);
空间复杂度为O(n),即申请了三个vector<int>,用于存放每次divide的元素。
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