74th LeetCode Weekly Contest Number of Subarrays with Bounded Maximum
We are given an array A
of positive integers, and two positive integers L
and R
(L <= R
).
Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L
and at most R
.
Example : Input: A = [2, 1, 4, 3] L = 2 R = 3 Output: 3 Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
Note:
- L, R and
A[i]
will be an integer in the range[0, 10^9]
. - The length of
A
will be in the range of[1, 50000]
.
有多少连续的子数组,必须存在[L,R],但不能超过R
应该知道了,A[i]在L,R之间是num+=(i-j),超过R需要记录j=i,但如何处理小于L的数字
因为必须存在[L,R],单纯的小于L不能算,那么就拿一个k记录一下不在[L,R]的位置,然后(i-j)-(i-k+1)就好了
1 class Solution { 2 public: 3 int numSubarrayBoundedMax(vector<int>& A, int L, int R) { 4 int aLen=A.size(); 5 int num=0; 6 int k=0; 7 int i=0,j=-1; 8 for(int i=0;i<aLen;i++){ 9 if(A[i]>R){ 10 k=i+1; 11 j=i; 12 }else if(A[i]<L){ 13 num+=(i-j)-(i-k+1); 14 }else{ 15 num+=(i-j); 16 k=i+1; 17 } 18 } 19 return num; 20 } 21 };