题目链接:LeetCode215
题解
第一种做法,直接排序完了取,时间复杂度O(logn)
class Solution {
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);
return nums[nums.length - k];
}
}
第二种做法,BFPRT算法,时间复杂度O(n)
class Solution {
public static int getMinKthByBFPRT(int[] arr,int k) {
int[] copyArr = new int[arr.length];
copyArr = copyArray(arr);
return bfprt(copyArr,0,copyArr.length - 1,k - 1);
}
public static int[] copyArray(int[] arr) {
int[] tmp = new int[arr.length];
for(int i = 0;i != arr.length;i++)
tmp[i] = arr[i];
return tmp;
}
public static int bfprt(int[] arr,int begin,int end,int i) {//begin到end范围内求第i小的数
if(begin == end)
return arr[begin];
int pivot = medianOfMedians(arr,begin,end);//中位数作为划分值
int[] pivotRange = partition(arr,begin,end,pivot);//进行划分,返回等于区域
if(i >= pivotRange[0] && i <= pivotRange[1])
return arr[i];
else if(i < pivotRange[0])
return bfprt(arr,begin,pivotRange[0] - 1,i);
else
return bfprt(arr,pivotRange[1] + 1,end,i);
}
public static int medianOfMedians(int[] arr,int begin,int end) {
int num = end - begin + 1;
int offset = num % 5 == 0 ? 0 : 1;
int[] mArr = new int[num / 5 + offset];
for(int i = 0; i < mArr.length;i++) {
int beginI = begin + i * 5;
int endI = beginI + 4;
mArr[i] = getMedian(arr,beginI,Math.min(end,endI));
}
return bfprt(mArr,0,mArr.length - 1,mArr.length / 2);
}
public static int getMedian(int[] arr,int begin,int end) {
Arrays.sort(arr,begin,end);
int sum = end + begin;
int mid = (sum / 2) + (sum % 2);
return arr[mid];
}
public static void insertSort(int[] arr,int begin,int end) {
for(int i = begin + 1;i != end + 1;i++) {
for(int j = i;j != begin;j--) {
if(arr[j - 1] > arr[j])
swap(arr,j - 1,j);
else
break;
}
}
}
public static int[] partition(int[] arr,int begin,int end,int pivotValue) {
int small = begin - 1;
int cur = begin;
int big = end + 1;
while(cur != big) {
if(arr[cur] < pivotValue)
swap(arr,++small,cur++);
else if(arr[cur] > pivotValue)
swap(arr,cur,--big);
else
cur++;
}
int[] range = new int[2];
range[0] = small + 1;
range[1] = big - 1;
return range;
}
public static void swap(int[] arr,int i,int j) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
}
public int findKthLargest(int[] nums, int k) {
return getMinKthByBFPRT(nums,nums.length - k + 1);
}
}