标签：nod 父节点 while better turn val version amp lang
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].
Time: O(NlogN), space: O(N) better than naive solution O(N^2)
public class Solution {
private class TreeNode {
public int val;
public int count = 1;
public TreeNode left, right;
public TreeNode(int val) {
this.val = val;
}
}
public List<Integer> countSmaller(int[] nums) {
List<Integer> res = new ArrayList<>();
if(nums == null || nums.length == 0) {
return res;
}
TreeNode root = new TreeNode(nums[nums.length - 1]);
res.add(0);
for(int i = nums.length - 2; i >= 0; i--) {
int count = addNode(root, nums[i]);
res.add(count);
}
Collections.reverse(res);
return res;
}
private int addNode(TreeNode root, int val) {
int curCount = 0;
while(true) {
if(val <= root.val) {
root.count++; // add the inversion count
if(root.left == null) {
root.left = new TreeNode(val);
break;
} else {
root = root.left;
}
} else {
curCount += root.count;
if(root.right == null) {
root.right = new TreeNode(val);
break;
} else {
root = root.right;
}
}
}
return curCount;
}
}
public class Solution {
int[] count;// 记录坐标的count
//记录排序后的坐标, 就是nums[index[i]] 为排序后的值
public List<Integer> countSmaller(int[] nums) {
List<Integer> ans = new ArrayList<>();
if (nums == null || nums.length == 0) {
return ans;
}
count = new int[nums.length];
int[] index = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
index[i] = i;
}
mergSort(nums, 0, nums.length - 1, index);
for (int i = 0; i < nums.length; i++) {
ans.add(count[i]);
}
return ans;
}
private void mergSort(int[] nums, int start, int end, int[] index) {
if (end <= start) return;
int mid = start + (end - start) / 2;
mergSort(nums, start, mid, index);
mergSort(nums, mid + 1, end, index);
merg(nums, start, end, index);
}
private void merg(int[] nums, int start, int end, int[] index) {
int mid = start + (end - start) / 2;
int left = start, right = mid + 1, newStart = 0, sum = 0;
int[] newIndex = new int[end - start + 1]; // 对坐标排序
while (left <= mid && right <= end) {
if (nums[index[right]] < nums[index[left]]) {
sum++;
newIndex[newStart++] = index[right++];
} else {
count[index[left]] += sum;
newIndex[newStart++] = index[left++];
}
}
while (left <= mid) {
count[index[left]] += sum;
newIndex[newStart++] = index[left++];
}
while (right <= end) {
newIndex[newStart++] = index[right++];
}
for (int i = start; i <= end; i++) {
index[i] = newIndex[i - start];
}
}
}
315. Count of Smaller Numbers After Self
标签：nod 父节点 while better turn val version amp lang
原文地址：http://www.cnblogs.com/apanda009/p/7298336.html
