leetcode 215. Kth Largest Element in an Array

Discription

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.

思路

快排思想的应用。
对于无序数组
$\{a_1, a_2, \cdots, a_n\}$
随机选取其中一个数$a_i$，按照 $> a_i$ 和 <=ai <script type="math/tex" id="MathJax-Element-969"><= a_i</script>将数组划分成两部分：
$\{a_{j_1}, a_{j_2}, \cdots, a_{j_{n_j}}\} > a_i >= \{a_{k_1}, a_{k_2}, \cdots, a_{k_{n_k}}\}$
如果$a_i$恰好为第k个元素，则$a_i$为所求解。否则进一步划分。

时间复杂度：$O(n)$
空间复杂度：$O(1)$

代码

class Solution {
public:
    int partition(vector<int> &nums, int start, int end)
    {
        int index = rand() % (end - start + 1) + start; //随机选择一个元素
        swap(nums[index], nums[end]);

        int big = start;    //下一个 >= nums[end] 的元素所应放置位置
        for (int i = start; i <= end; i++)
        {
            if (nums[i] > nums[end])        // 注意: >= 无法通过 {1}， k = 1 样例，越界错误
            {
                if (i != big)
                    swap(nums[i], nums[big]);

                big++;
            }
        }
        swap(nums[end], nums[big]);
        return big;
    }

    int findKthLargest(vector<int>& nums, int k) {
        int left = 0, right = nums.size() - 1;

        while (true)
        {
            int loc = partition(nums, left, right);
            if (loc == k - 1)
                return nums[loc];

            if (loc < k - 1)
                left = loc + 1;
            else
                right = loc - 1;
        }
    }
};

特殊样例

{1}， 1

扩展

该方法时间复杂度为什么是$O(n)$?