【Leetcode】Range Sum Query - Mutable

题目链接：https://leetcode.com/problems/range-sum-query-mutable/

题目：

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
The update(i,
val) function modifies nums by
updating the element at index i to val.

Example:

Given nums = [1, 3, 5]

sumRange(0, 2) -> 9
update(1, 2)
sumRange(0, 2) -> 8

Note:

The array is only modifiable by the update function.
You may assume the number of calls to update and sumRange function is distributed evenly.

思路：
segment tree基础，创建tree用分治，求sum要注意所求的区间和节点区间的关系。
算法：
public class NumArray {

Node root = null;

public NumArray(int[] nums) {
root = create(nums, 0, nums.length - 1);
}

void update(int i, int val) {
update(root, i, val);
}

public int sumRange(int i, int j) {
return sum(root, i, j);
}

public class Node {
int lidx;
int ridx;
int val;
Node lchild;
Node rchild;

public Node(int lidx, int ridx, int val) {
this.lidx = lidx;
this.ridx = ridx;
this.val = val;
}

}

public Node create(int[] nums, int l, int r) {
if (l == r)
return new Node(l, r, nums[r]);
Node lchild = create(nums, l, (l + r) / 2);
Node rchild = create(nums, (l + r) / 2 + 1, r);
Node root = new Node(lchild.lidx, rchild.ridx, lchild.val + rchild.val);
root.lchild = lchild;
root.rchild = rchild;
return root;
}

public int sum(Node root, int l, int r) {
if (root.ridx < l || root.lidx > r)// 所求区间与节点区间不相交
return 0;

if (l <= root.lidx && root.ridx <= r)// 所求区间包含了 节点区间
return root.val;

return sum(root.lchild, l, r) + sum(root.rchild, l, r);//所求区间和节点区间相交
}

int diff = 0;// 需要从index叶节点 往上更新的值
public void update(Node root, int index, int newVal) {
if (root.lidx == root.ridx && index == root.lidx) {// 要更新的叶节点
diff = newVal - root.val;
root.val = newVal;
return;
}
if (index <= (root.lidx + root.ridx) / 2)
update(root.lchild, index, newVal);
else
update(root.rchild, index, newVal);

root.val += diff;
}
}