先想到暴力解法直接用数组保留前i项和,求sumRange可以直接做差,但是发现会超时。
后来想到分块,和暴力求法类似,AC。
class NumArray {
public:
vector<int> sums;
vector<int> &nums;
int s;
NumArray(vector<int>& nums) : nums(nums){
int n = nums.size();
s = sqrt(n);
sums.resize((n + s - 1) / s);
for(int i = 0; i < n; i ++){
sums[i/s] += nums[i];
}
}
void update(int index, int val) {
sums[index / s] += val - nums[index];
nums[index] = val;
}
int sumRange(int left, int right) {
int l1 = left / s, i1 = left % s, r2 = right / s, i2 = right % s;
if(l1 == r2){
return accumulate(nums.begin() + l1 * s + i1, nums.begin() + l1 * s + i2 + 1, 0);
}
int sum1 = accumulate(nums.begin() + l1 * s + i1, nums.begin() + l1 * s + s, 0);
int sum2 = accumulate(nums.begin() + r2 * s, nums.begin() + r2 * s + i2 + 1, 0);
int sum3 = accumulate(sums.begin() + l1 + 1, sums.begin() + r2, 0);
return sum1 + sum2 + sum3;
}
};
/**
* 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);
*/
参考了题解知道了accumulate求和函数。
又看到了线段树的解法
class NumArray {
private:
vector<int> segmentTree;
int n;
void build(int node, int s, int e, vector<int> &nums) {
if (s == e) {
segmentTree[node] = nums[s];
return;
}
int m = s + (e - s) / 2;
build(node * 2 + 1, s, m, nums);
build(node * 2 + 2, m + 1, e, nums);
segmentTree[node] = segmentTree[node * 2 + 1] + segmentTree[node * 2 + 2];
}
void change(int index, int val, int node, int s, int e) {
if (s == e) {
segmentTree[node] = val;
return;
}
int m = s + (e - s) / 2;
if (index <= m) {
change(index, val, node * 2 + 1, s, m);
} else {
change(index, val, node * 2 + 2, m + 1, e);
}
segmentTree[node] = segmentTree[node * 2 + 1] + segmentTree[node * 2 + 2];
}
int range(int left, int right, int node, int s, int e) {
if (left == s && right == e) {
return segmentTree[node];
}
int m = s + (e - s) / 2;
if (right <= m) {
return range(left, right, node * 2 + 1, s, m);
} else if (left > m) {
return range(left, right, node * 2 + 2, m + 1, e);
} else {
return range(left, m, node * 2 + 1, s, m) + range(m + 1, right, node * 2 + 2, m + 1, e);
}
}
public:
NumArray(vector<int>& nums) : n(nums.size()), segmentTree(nums.size() * 4) {
build(0, 0, n - 1, nums);
}
void update(int index, int val) {
change(index, val, 0, 0, n - 1);
}
int sumRange(int left, int right) {
return range(left, right, 0, 0, n - 1);
}
};
作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/range-sum-query-mutable/solution/qu-yu-he-jian-suo-shu-zu-ke-xiu-gai-by-l-76xj/
来源:力扣(LeetCode)
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