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CountofSmallerNumbersAfterSelf.java
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package leetcode;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
* Project Name : Leetcode
* Package Name : leetcode
* File Name : CountofSmallerNumbersAfterSelf
* Creator : Edward
* Date : Nov, 2017
* Description : 315. Count of Smaller Numbers After Self
*/
public class CountofSmallerNumbersAfterSelf {
/**
* You are given an integer array nums and you have to return a new counts array.
* The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
Example:
Given nums = [5, 2, 6, 1]
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].
[5, 2, 6, 1]
int[] Arrays.asList()
0 1 1 2
time : O(n^2)
space : O(n)
*/
public List<Integer> countSmaller(int[] nums) {
Integer[] res = new Integer[nums.length];
List<Integer> list = new ArrayList<>();
for (int i = nums.length - 1; i >= 0; i--) {
int index = findIndex(list, nums[i]);
res[i] = index;
list.add(index, nums[i]);
}
return Arrays.asList(res);
}
private int findIndex(List<Integer> list, int target) {
if (list.size() == 0) return 0;
int start = 0;
int end = list.size() - 1;
if (list.get(end) < target) return end + 1;
if (list.get(start) >= target) return 0;
while (start + 1 < end) {
int mid = (end - start) / 2 + start;
if (list.get(mid) < target) {
start = mid + 1;
} else {
end = mid;
}
}
if (list.get(start) >= target) return start;
return end;
}
}