LeetCode Count of Range Sum

原题链接在这里：https://leetcode.com/problems/count-of-range-sum/

题目：

Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive.

Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.

Example:
Given nums = [-2, 5, -1], lower = -2, upper = 2,
Return 3.
The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

题解：

题目的意思是说给了一个int array, 计算有多少subarray的sum在[lower, upper]区间内. 给的例子是index.

建立BST，每个TreeNode的val是prefix sum. 为了避免重复的TreeNode.val, 设置一个count记录多少个重复TreeNode.val, 维护leftSize, 记录比该节点value小的节点个数，rightSize同理.

由于RangeSum S(i,j)在[lower,upper]之间的条件是lower<=sums[j+1]-sums[i]<=upper. 所以我们每次insert一个新的PrefixSum sums[k]进这个BST之前，先寻找一下rangeSize该BST内已经有多少个PrefixSum, 叫它sums[t]吧, 满足lower<=sums[k]-sums[t]<=upper, 即寻找有多少个sums[t]满足:

sums[k]-upper<=sums[t]<=sums[k]-lower

BST提供了countSmaller和countLarger的功能，计算比sums[k]-upper小的RangeSum数目和比sums[k]-lower大的数目，再从总数里面减去，就是所求

Time Complexity: O(nlogn). Space: O(n).

AC Java:

 public class Solution {
     public int countRangeSum(int[] nums, int lower, int upper) {
         if(nums == null || nums.length == 0){
             return 0;
         }
         int res = 0;
         long [] sum = new long[nums.length+1];
         for(int i = 1; i<sum.length; i++){
             sum[i] = sum[i-1] + nums[i-1];
         }

         TreeNode root = new TreeNode(sum[0]);
         for(int i = 1; i<sum.length; i++){
             res += rangeSize(root, sum[i]-upper, sum[i]-lower);
             insert(root, sum[i]);
         }
         return res;
     }

     private TreeNode insert(TreeNode root, long val){
         if(root == null){
             return new TreeNode(val);
         }
         if(root.val == val){
             root.count++;
         }else if(root.val > val){
             root.leftSize++;
             root.left = insert(root.left, val);
         }else if(root.val < val){
             root.rightSize++;
             root.right = insert(root.right, val);
         }
         return root;
     }

     private int countSmaller(TreeNode root, long val){
         if(root == null){
             return 0;
         }
         if(root.val == val){
             return root.leftSize;
         }else if(root.val > val){
             return countSmaller(root.left, val);
         }else{
             return root.leftSize + root.count + countSmaller(root.right, val);
         }
     }

     private int countLarget(TreeNode root, long val){
         if(root == null){
             return 0;
         }
         if(root.val == val){
             return root.rightSize;
         }else if(root.val > val){
             return countLarget(root.left, val) + root.count + root.rightSize;
         }else{
             return countLarget(root.right, val);
         }
     }

     private int rangeSize(TreeNode root, long lower, long upper){
         int total = root.leftSize + root.count + root.rightSize;
         int smaller = countSmaller(root, lower);
         int larger = countLarget(root, upper);
         return total - smaller - larger;
     }
 }

 class TreeNode{
     long val;
     int count;
     int leftSize;
     int rightSize;
     TreeNode left;
     TreeNode right;
     public TreeNode(long val){
         this.val = val;
         this.count = 1;
         this.leftSize = 0;
         this.rightSize = 0;
     }
 }

Reference: http://www.cnblogs.com/EdwardLiu/p/5138198.html