LintCode 5 / LeetCode 215 Kth Largest Element

思路

quick select / partition（二者是一个东西）

复杂度

时间复杂度O(n)
空间复杂度O(n)

代码

public class Solution {
    /*** @param n: An integer* @param nums: An array* @return: the Kth largest element*/public int kthLargestElement(int n, int[] nums) {
    // write your code hereif (nums == null) {
    return -1;}return quickSelect(nums, 0, nums.length - 1, n);}private int quickSelect(int[] nums, int start, int end, int k) {
    if (start >= end) {
    return nums[start];}int left = start, right = end;int pivot = nums[(left + right) / 2];while (left <= right) {
    while (left <= right && nums[left] > pivot) {
    left++;}while (left <= right && nums[right] < pivot) {
    right--;}if (left <= right) {
    int t = nums[left];nums[left] = nums[right];nums[right] = t;left++;right--;}}if (start + (k - 1) <= right) {
    return quickSelect(nums, start, right, k);}if (start + (k - 1) >= left) {
    return quickSelect(nums, left, end, k - (left - start));}return nums[left - 1];}
}