"215-kth largest Element in an array (number of k large in array)"
"leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index"
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Original Question
Find the kth largest element in an unsorted array. Note that it was the kth largest element in the sorted order and not the kth distinct element.
For example,
Given [3,2,1,5,6,4]
and k = 2, return 5.
Note:
You may assume k are always valid, 1≤k≤array ' s length.
Main Topic
Find the element of the K-Large from an unordered array. Note that the K-large after sorting, and not the K-non-repeating element can assume that K must be valid, 1≤k≤ array length
Thinking of solving problems
O (n) Solution: Fast selection (Quickselect) algorithm
Code Implementation
Algorithm implementation class
Import java.util.Collections; Public classSolution { Public int findkthlargest(int[] Nums,intK) {if(K <1|| Nums = =NULL|| Nums.length < K) {Throw NewIllegalArgumentException (); }returnFindkthlargest (Nums,0, Nums.length-1, k); } Public int findkthlargest(int[] Nums,intStartintEndintK) {//Center value intPivot = Nums[start];intLo = start;inthi = end; while(Lo < HI) {//Move the number less than the center value to the left of the array while(Lo < hi && Nums[hi] >= pivot) {hi--; } Nums[lo] = Nums[hi];//Move the number greater than the center value to the right of the array while(Lo < hi && Nums[lo] <= pivot) {lo++; } Nums[hi] = Nums[lo]; } Nums[lo] = pivot;//If you have found if(End-lo +1= = k) {returnPivot }//K-large number at the right of Lo position Else if(End-lo +1> k) {returnFindkthlargest (Nums, lo +1, end, K); }//K-large number at the left of the LO position Else{//K (end-lo+1) //(END-LO+1): Indicates the number of elements from the LO position to the end position, or the right half. //The original K-Large becomes K (end-lo+1) Large returnFindkthlargest (nums, Start, lo-1, K-(End-lo +1)); } }}
Evaluation Results
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Special Instructions
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"Leetcode-Interview algorithm classic-java implementation" "215-kth largest Element in an array (number of K in array)"