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/**

* 这题可以用segment tree或binary index tree(bit)，两者时间复杂度都是lgn，不过segment tree因为用到树，

* 而bit用的是数组，所以bit更省空间。

*

* 关于BIT，可以参考http://www.geeksforgeeks.org/binary-indexed-tree-or-fenwick-tree-2/

* 关于Segment Tree，可以参考 https://discuss.leetcode.com/topic/29918/17-ms-java-solution-with-segment-tree

*

* https://leetcode.com/articles/range-sum-query-mutable/

*/

public class NumArrayII {

private int[] bit;

private int[] nums;

public NumArrayII(int[] nums) {

this.nums = nums;

bit = new int[nums.length + 1];

for (int i = 0; i < nums.length; i++) {

init(i, nums[i]);

}

}

public void update(int i, int val) {

int diff = val - nums[i];

nums[i] = val;

init(i, diff);

}

private void init(int i, int val) {

i++;

while (i < bit.length) {

bit[i] += val;

i += i & (-i);

}

}

private int getSum(int i) {

i++;

int sum = 0;

while (i > 0) {

sum += bit[i];

i -= i & (-i);

}

return sum;

}

public int sumRange(int i, int j) {

return getSum(j) - getSum(i - 1);

}

/**

private SegmentTreeNode mRoot;

public NumArrayII(int[] nums) {

mRoot = buildTree(nums, 0, nums.length - 1);

}

private SegmentTreeNode buildTree(int[] nums, int start, int end) {

if (start > end) {

return null;

}

SegmentTreeNode root = new SegmentTreeNode(start, end);

if (start == end) {

root.sum = nums[start];

return root;

}

int mid = start + ((end - start) >>> 1);

root.left = buildTree(nums, start, mid);

root.right = buildTree(nums, mid + 1, end);

root.sum = root.left.sum + root.right.sum;

return root;

}

public void update(int i, int val) {

update(mRoot, i, val);

}

private void update(SegmentTreeNode root, int i, int val) {

int start = root.start, end = root.end;

if (start == end) {

root.sum = val;

return;

}

int mid = start + ((end - start) >>> 1);

if (i <= mid) {

update(root.left, i, val);

} else {

update(root.right, i, val);

}

root.sum = root.left.sum + root.right.sum;

}

public int sumRange(int i, int j) {

return sumRange(mRoot, i, j);

}

public int sumRange(SegmentTreeNode root, int i, int j) {

int start = root.start, end = root.end;

if (start == i && end == j) {

return root.sum;

}

int mid = start + ((end - start) >>> 1);

if (j <= mid) {

return sumRange(root.left, i, j);

} else if (i > mid) {

return sumRange(root.right, i, j);

} else {

return sumRange(root.left, i, mid) + sumRange(root.right, mid + 1, j);

}

}*/

}

// Your NumArray object will be instantiated and called as such:

// NumArray numArray = new NumArray(nums);

// numArray.sumRange(0, 1);

// numArray.update(1, 10);

// numArray.sumRange(1, 2);