You are given an array of integers nums and a sliding window of size k
which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. You have to compute the maximum inside the window.
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
class Solution {
public:
vector<int> maxSlidingWindow(vector<int> &nums, int k) {
int n=nums.size();
vector<int> answer;
for(int i=0; i<n; i ){
int mx = INT_MIN;
for(int j=1; j<i k; j ){
mx = max(mx, nums[j]);
}
answer.push_back(mx);
}
while(answer.size()>n-k 1){
answer.pop_back();
}
return answer;
}
};
But throws an error
=================================================================
==31==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x603000000090 at pc 0x000000345e1e bp 0x7ffef610bff0 sp 0x7ffef610bfe8
READ of size 4 at 0x603000000090 thread T0
#2 0x7fb1bb36d0b2 (/lib/x86_64-linux-gnu/libc.so.6 0x270b2)
0x603000000090 is located 0 bytes to the right of 32-byte region [0x603000000070,0x603000000090)
allocated by thread T0 here:
#6 0x7fb1bb36d0b2 (/lib/x86_64-linux-gnu/libc.so.6 0x270b2)
Shadow bytes around the buggy address:
0x0c067fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c067fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c067fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c067fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c067fff8000: fa fa 00 00 00 07 fa fa fd fd fd fa fa fa 00 00
=>0x0c067fff8010: 00 00[fa]fa 00 00 00 00 fa fa fa fa fa fa fa fa
0x0c067fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff8060: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
Shadow gap: cc
==31==ABORTING
CodePudding user response:
for(int i=0; i<n; i ){
int mx = INT_MIN;
for(int j=1; j<i k; j ){
mx = max(mx, nums[j]);
Say k
is 3, and n
is 2. When i
= 1, j
will go to 3. nums[3]
is then out-of-bounds. You need to carefully review j=1
and j<i k
. Both are wrong.
Also, int mx = nums[i];
is more elegant and doesn't involve special constants like INT_MIN
.
CodePudding user response:
Using the deque did the trick
vector<int> maxSlidingWindow(vector<int> &nums, int k) {
int n = nums.size();
deque<int> dq(k);
vector<int> answer;
for (int i = 0; i < k; i ) {
while (dq.size() && nums[i] >= nums[dq.back()]) {
dq.pop_back();
}
dq.push_back(i);
}
for (int i = k; i < n; i ){
answer.push_back(nums[dq.front()]);
while(dq.size() && dq.front() <= i - k) {
dq.pop_front();
}
while(dq.size() && nums[i] >= nums[dq.back()]) {
dq.pop_back();
}
dq.push_back(i);
}
answer.push_back(nums[dq.front()]);
return answer;
}