leetcode---Range Sum Query - Mutable---二叉索引树

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.

class NumArray
{
public:
vector<int> c;
vector<int> num;

NumArray(vector<int> nums)
{
num = nums;
int n = nums.size();
c.resize(n+1);
for(int i=0; i<n; i++)
add(i+1, nums[i]);
}

int lowbit(int p)
{
return p & -p;
}

void add(int i, int num)
{
while(i < c.size())
{
c[i] += num;
i += lowbit(i);
}
}

int sum(int i)
{
int r = 0;
while(i > 0)
{
r += c[i];
i -= lowbit(i);
}
return r;
}

void update(int i, int val)
{
int delta = val - num[i];
num[i] = val;
add(i+1, delta);
}

int sumRange(int i, int j)
{
return sum(j+1) - sum(i);
}
};

/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/

参考

http://blog.csdn.net/u010270082/article/details/36506033