题目原文:
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.
题目大意:
寻找一个数组中第k大的数。
题目分析:
(1) 朴素解法O(nlogn):排序,然后返回倒数第k个元素。
(2) Quick Select算法O(n):
S1.选择一个中轴(可以使用快排中的三者取中法),比中轴大的数放到左边,比中轴小的数放到右边;
S2.然后求出左边的长度l,若l==k,则中轴即为所求;若l>k,则从左边数组里面找第k大的数,若l
public class Solution {
public int findKthLargest(int[] nums, int k) {
return select(nums, k-1);
}
private int select(int[] nums, int k) {
int left = 0, right = nums.length-1;
while(true) {
if(left == right)
return nums[left];
int pivotIndex = medianOf3(nums, left, right);
pivotIndex = partition(nums, left, right, pivotIndex);
if(pivotIndex == k)
return nums[k];
else if(pivotIndex > k)
right = pivotIndex-1;
else
left = pivotIndex+1;
}
}
//Use median-of-three strategy to choose pivot
private int medianOf3(int[] nums, int left, int right) {
int mid = left + (right - left) / 2;
if(nums[right] > nums[left])
swap(nums, left, right);
if(nums[right] > nums[mid])
swap(nums, right, mid);
if(nums[mid] > nums[left])
swap(nums,left, mid);
return mid;
}
private int partition(int[] nums, int left, int right, int pivotIndex) {
int pivotValue = nums[pivotIndex];
swap(nums, pivotIndex, right);
int index = left;
for(int i = left; i < right; ++i) {
if(nums[i] > pivotValue) {
swap(nums, index, i);
++index;
}
}
swap(nums, right, index);
return index;
}
private void swap(int[] nums, int a, int b) {
int temp = nums[a];
nums[a] = nums[b];
nums[b] = temp;
}
}
成绩:
方法一:7ms,beats 74.67%,众数4ms,13.67%
方法二:2ms,beats 97.12%
Cmershen的碎碎念:
这道题好像是一道很经典的题,似乎在《算法导论》中有对这道题大篇幅的详细描述。