力扣每日一题

力扣每日一题

307. 区域和检索 - 数组可修改 - 力扣（LeetCode）

看题面单点修改加区间查询 == 一眼线段树 >> 由于没有复杂修改操作 - > 简化为树状数组求解

int init = []()
{cin.tie(0) -> sync_with_stdio(false);return 0;
}();class NumArray {
public:NumArray(vector<int>& nums) : sum(nums.size() + 1, 0), num(nums), n(nums.size()) {for(int i = 0;i < n;i++)add(i + 1,nums[i]);}int lowbit(int x){return x & -x;}void add(int pos, int val){for(; pos <= n; pos += lowbit(pos))sum[pos] += val;}void update(int index, int val) {add(index + 1, val - num[index]);num[index] = val;}int ask(int x){int ans = 0;for(; x > 0; x -= lowbit(x))ans += sum[x];return ans;}int sumRange(int left, int right) {return ask(right + 1) - ask(left);}
private:
vector<int>sum;
vector<int>&num;
int n;
};/*** Your NumArray object will be instantiated and called as such:* NumArray* obj = new NumArray(nums);* obj->update(index,val);* int param_2 = obj->sumRange(left,right);*/

还有一个有意思的解法

int init = []()
{cin.tie(0) -> sync_with_stdio(false);return 0;
}();class NumArray 
{
public:NumArray(vector<int>& nums) {n = nums.size();v.resize(n << 1);for(int i = n; i < (n << 1); i++)v[i] = nums[i - n];for(int i = n - 1; i > 0; i--)v[i] = v[i << 1] + v[(i << 1) + 1];}void update(int index, int val) {int i = n + index;int diff = val - v[i];v[i] = val;while(i > 1){v[i >> 1] += diff;i >>= 1;}}int sumRange(int left, int right) {left += n;right += n; int sum = 0;while(left <= right){if(left & 1){ sum += v[left]; left++; }if(!(right & 1)){ sum += v[right]; right--; }  left >>= 1;right >>= 1;}return sum;}
private:vector<int> v;int n;
};