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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.
这道没啥可说的。最容易想到的办法就是不管大小先把整个数列sorting，然后直接return第（数列.length-k）个数即可。但是整个如果数列很长的话就比较浪费时间啦。代码很简单。如下。
public class Solution {
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);
return nums[nums.length-k];
}
}
但是啦，个人觉得其实最快的应该是用quick select来做。当然了代码就比sorting要略麻烦一些。
qucike select是直接用前后开始找，就不管这个数列多长，速度都不会有太大影响。
public class Solution {
public int findKthLargest(int[] nums, int k) {
int start = 0, end = nums.length - 1, index = nums.length - k;
while (start < end) {
int pivot = getKth(nums, start, end);
if (pivot < index) {
start = pivot + 1;
}else if (pivot > index) {
end = pivot - 1;
}else{
return nums[pivot];
}
}
return nums[start];
}
private int getKth(int[] nums, int start, int end) {
int pivot = start, temp;
while (start <= end) {
while (start <= end && nums[start] <= nums[pivot]) {
start++;
}
while (start <= end && nums[end] > nums[pivot]){
end--;
}
if (start > end) {
break;
}
temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
}
temp = nums[end];
nums[end] = nums[pivot];
nums[pivot] = temp;
return end;
}
}
[LeetCode] Kth Largest Element in an Array
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原文地址：http://www.cnblogs.com/orangeme404/p/4726577.html
