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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:
题解：树状数组最基本的用法（点更新，区间查询）。需要注意的是这里的点更新不是把点加一个值，而是完全更新一个点的值，这就需要不仅仅更新树状数组，原数组的值也应该更新，因为可能下一个还会更新这个点，那么就需要知道它之前的值是多少。
class NumArray { public: NumArray(vector<int> &nums) { n = nums.size(); c=vector<int>(n+1,0); a=vector<int>(n+1,0); for(int i=1;i<=n;i++){ update(i-1,nums[i-1]); } } int lowbit(int x){ return x&-x; } void update(int i, int val) { int old = a[i+1]; for(int k=i+1;k<=n;k+=lowbit(k)){ c[k]-=old; c[k]+=val; } a[i+1]=val; } int sum(int x){ int s=0; while(x>0){ s+=c[x]; x-=lowbit(x); } return s; } int sumRange(int i, int j) { return sum(j+1)-sum(i); } private: int n; vector<int>c; vector<int>a; }; // Your NumArray object will be instantiated and called as such: // NumArray numArray(nums); // numArray.sumRange(0, 1); // numArray.update(1, 10); // numArray.sumRange(1, 2);
leetcode 307. Range Sum Query - Mutable(树状数组)
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原文地址：http://www.cnblogs.com/zywscq/p/5410994.html
