307 Range Sum Query - Mutable
阿新 • • 發佈:2018-04-14
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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:
The array is only modifiable by the update function.
You may assume the number of calls to update and sumRange function is distributed evenly.
詳見:https://leetcode.com/problems/range-sum-query-mutable/description/
C++:
class NumArray { public: NumArray(vector<int> nums) { num.resize(nums.size()+1); bit.resize(nums.size()+1); for(int i=0;i<nums.size();++i) { update(i,nums[i]); } } void update(int i, int val) { int diff=val-num[i+1]; for(int j=i+1;j<num.size();j+=(j&-j)) { bit[j]+=diff; } num[i+1]=val; } int sumRange(int i, int j) { return getSum(j+1)-getSum(i); } int getSum(int i) { int res=0; for(int j=i;j>0;j-=(j&-j)) { res+=bit[j]; } return res; } private: vector<int> num; vector<int> bit; }; /** * 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); */
參考:https://www.cnblogs.com/grandyang/p/4985506.html
307 Range Sum Query - Mutable