LeetCode *** 307. Range Sum Query - Mutable (Binary Indexed Trees)

题目：

Given an integer array nums, find the sum of the elements between indicesi and
j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at indexi 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.

分析：

代码：

class NumArray {
public:
vector<int> record, tree;
NumArray(vector<int> &nums) {
int size = nums.size(), idx;
record = nums;
record.insert(record.begin(), 0);
tree = vector<int>(size + 1, 0);

for (int i = 1; i <= size; ++i) {
idx = i - (i&-i) + 1;
for (int j = idx; j <= i; ++j)
tree[i] += record[j];
}
}

void update(int i, int val) {
int differ = val- record[i + 1], idx = i + 1;
record[i + 1] = val;
while (idx<record.size()) {
tree[idx] += differ;
idx += (idx&-idx);
}
}

int sumRange(int i, int j) {
int sum1 = 0,sum2=0, idx = i,jdx=j+1;

while (idx>0||jdx>0) {
if (idx > 0) {
sum1 += tree[idx];
idx -= (idx&-idx);
}
if (jdx > 0) {
sum2 += tree[jdx];
jdx -= (jdx&-jdx);
}
}
return sum2-sum1;
}
};