LintCode 207. 区间求和 II（线段树）

1. 题目

在类的构造函数中给一个整数数组, 实现两个方法 query(start, end) 和 modify(index, value):

对于 query(start, end), 返回数组中下标 start 到 end 的和。
对于 modify(index, value), 修改数组中下标为 index 上的数为 value.

样例1
输入:
[1,2,7,8,5]
[query(0,2),modify(0,4),query(0,1),modify(2,1),query(2,4)]
输出: [10,6,14]
说明:
给定数组 A = [1,2,7,8,5].
在query(0, 2)后, 1 + 2 + 7 = 10,
在modify(0, 4)后, 将 A[0] 修改为 4， A = [4,2,7,8,5].
在query(0, 1)后, 4 + 2 = 6.
在modify(2, 1)后, 将 A[2] 修改为 1，A = [4,2,1,8,5].
After query(2, 4), 1 + 8 + 5 = 14.样例2
输入:
[1,2,3,4,5]
[query(0,0),query(1,2),quert(3,4)]
输出: [1,5,9]
说明:
1 = 1
2 + 3 = 5
4 + 5 = 9挑战
query 和 modify 的时间复杂度需要为O(logN).

2. 解题

参考线段树 Segment Tree

class node
{
public:int sum;int start, end;node *left, *right;node(int s, int e, int v):start(s),end(e),sum(v){left = right = NULL;}static node* build(vector<int>& A, int l, int r){if(l > r)return NULL;node* head = new node(l,r,A[l]);if(l == r)return head;int mid = l+((r-l)>>1);head->left = build(A,l,mid);head->right = build(A,mid+1,r);head->sum = 0;if(head->left)head->sum += head->left->sum;if(head->right)head->sum += head->right->sum;return head;}static long long query(node* head, int s, int e){if(s > head->end || e < head->start)return 0;if(head->start >= s && head->end <= e)return head->sum;int vl = query(head->left, s, e);int vr = query(head->right,s, e);return vl+vr;}static void modify(node* head, int id, int val){if(head->start == head->end){head->sum = val;return;}int mid = (head->start + head->end)/2;if(id > mid)modify(head->right, id, val);elsemodify(head->left, id, val);head->sum = 0;if(head->left)head->sum += head->left->sum;if(head->right)head->sum += head->right->sum;}
};
class Solution {node *head;
public:Solution(vector<int> A) {head = node::build(A,0,A.size()-1);}long long query(int start, int end) {return node::query(head, start,end);}void modify(int index, int value) {node::modify(head, index,value);}
};