leetcode 215. Kth Largest Element in an Array 寻找第k个大的数---------- java

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.

 

寻找一个数组中第k大的数。

 

1、建立一个大小为k的数组，每次维护这个数组，属于暴力做法。

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        int len = nums.length;
        int[] result = new int[k];
        for (int i = 0; i < k ; i++){
            result[i] = nums[i];
        }
        for (int i = k; i < len; i++){
            int min = result[0];
            int pos = 0;
            for (int j = 1; j < k; j++){
                if (min > result[j]){
                    min = result[j];
                    pos = j;
                }
            }
            if (nums[i] > min){
                result[pos] = nums[i];
            }
        }
        int ans = result[0];
        for (int i = 1; i < k; i++){
            System.out.println(result[i] + " " + result[0]);
            ans = Math.min(ans, result[i]);
        }
        return ans;
    }
}

2、利用快排

 

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        return QuickSort(nums, k, 0, nums.length - 1);
    }
    
    public int QuickSort(int[] nums, int k, int start, int end){
        int left = start, right = end;
        int num = nums[start];
        for (int i = end; i > start; i--){
            if (nums[i] >= num){
                swap(nums, end--, i);
            }
        }
        swap(nums, end, start);
        if (k == nums.length - end){
            return nums[end];
        } else if (k > nums.length - end){
            return QuickSort(nums, k, left, end - 1);
        } else {
            return QuickSort(nums, k, end + 1, right);
        }
    }
    public void swap(int[] nums, int a, int b){
        int num = nums[a];
        nums[a] = nums[b];
        nums[b] = num;
    }
}

 

 

3、discuss中的快排。速度更快

public class Solution {
    
    public int findKthLargest(int[] nums, int k) {

        return select(nums, k-1);
    }
    
    // Quick select
    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;
    }

}