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