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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
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Solution:
用values和sums分别记录nums和nums的前n项和，update的时候，sums中index大于i的都要变化。
1 class NumArray 2 { 3 public: 4 NumArray(vector<int> &nums) 5 { 6 sums.push_back(0); 7 for (int i = 0; i < nums.size(); i++) 8 { 9 sums.push_back(sums[i] + nums[i]); 10 values.push_back(nums[i]); 11 } 12 } 13 14 void update(int i, int val) 15 { 16 int diff = val - values[i]; 17 for (int k = i + 1; k < sums.size(); k++) 18 sums[k] = sums[k] + diff; 19 values[i] = val; 20 } 21 22 int sumRange(int i, int j) 23 { 24 return sums[j + 1] - sums[i]; 25 } 26 private: 27 vector<int> sums; 28 vector<int> values; 29 };
[leetcode] 307. Range Sum Query - Mutable
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原文地址：http://www.cnblogs.com/ym65536/p/5722480.html
