LintCode_205 Interval Minimum Number

Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers 

[start,
end]

. For each query, calculate the minimum number between index start and end in the given array, return the result list.


 Notice

We suggest you finish problem Segment
Tree Build, Segment Tree Query and Segment
Tree Modify first.

Have you met this question in a real interview? 

Yes

Example

For array 

[1,2,7,8,5]

,
and queries 

[(1,2),(0,4),(2,4)]

, return 

[2,1,5]

Challenge 

O(logN) time for each query

线段树实现：

/**
* Definition of Interval:
* classs Interval {
* int start, end;
* Interval(int start, int end) {
* this->start = start;
* this->end = end;
* }
*/
class Solution {
public:
/**
*@param A, queries: Given an integer array and an query list
*@return: The result list
*/
vector<int> intervalMinNumber(vector<int> &A, vector<Interval> &queries) {
// write your code here
vector<int> res;
vector<int> dat = init(A);
for (int i = 0; i < queries.size(); i++) {
int min_val = query(dat, queries[i].start, queries[i].end + 1, 0, 0, dat.size() / 2);
//注意区间是左闭右开；
res.push_back(min_val);
}
return res;

}
vector<int> init (vector<int> &A) {
int len = A.size();
int n = 1;
while ( n <= len) {
n*=2;
}
vector<int> dat(2 * n, INT_MAX);
for (int i = 0; i < A.size(); i++) {
update(dat, i, A[i]);
}
// for (int i = 0; i < dat.size(); i++) {
// cout<<dat[i]<<" ";
// }
return dat;
}
void update(vector<int> &dat, int k, int value) {
int n = dat.size() / 2;
k += n - 1; //注意要去掉第一个；
dat[k] = value;
while (k > 0) {
k = (k - 1) / 2;
dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
}
}
int query (vector<int> &dat, int a, int b, int k, int l, int r) {
// l 和 r 代表节点k所代表的区间
//注意区间是左闭右开；
if (r <= a || b <= l) {
return INT_MAX;
}
if (a <= l && r <= b) {
return dat[k];
}else {
int v1 = query(dat, a, b, k * 2 + 1, l, (l + r) / 2);
int v2 = query(dat, a, b, k * 2 + 2, (l + r) / 2, r);
return min(v1, v2);
}
}
};