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All solutions to the problems in Leetcode will be coded in C99 or C++11 enclosed with critical comments.
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All solutions to the problems in Leetcode will be coded in C99 or C++11 enclosed with critical comments. And all the solutions will be as much diversified as possible while meantime keep their efficiency and conciseness.


All the solutions will be arranged as the following way displaying the author, e-mail, working time, problem description, the URL of the problem and maybe some brilliant references and then the code along with performance and detailed comments for the problem.

Author      : LHearen
E-mail      :
Time        : 2016-03-11 22:47
Description : Given an array nums, there is 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.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
1 [3  -1  -3] 5  3  6  7       3
1  3 [-1  -3  5] 3  6  7       5
1  3  -1 [-3  5  3] 6  7       5
1  3  -1  -3 [5  3  6] 7       6
1  3  -1  -3  5 [3  6  7]      7
Therefore, return the max sliding window as [3,3,5,5,6,7].
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.
Follow up:
Could you solve it in linear time?
Source      :
Reference   :
//AC - 40ms;
int* maxSlidingWindow(int* nums, int size, int k, int* returnSize)
    *returnSize = 0;
    int* arr = (int*)malloc(sizeof(int)*(size-k+1)); //store the result;
    int* queue = (int*)malloc(sizeof(int)*size); //record the index;
    int begin=0, end=-1;
    for(int i = 0; i < size; i++)
        if(end-begin>-1 && queue[begin]==i-k) begin++; //remove the index that is out of range k;
        while(end-begin>-1 && nums[queue[end]]<nums[i]) end--; //remove the smallers in the range k ensuring the queue[begin] points to the biggest in the range k;
        queue[++end] = i;
        if(i >= k-1) //only after k elements, we can start to collect;
            *returnSize += 1;
            arr[*returnSize-1] = nums[queue[begin]]; //the previous two removing operations will ensure the queue[begin] indexing the biggest in the k-range;
    return arr;


There are different folders containing the solutions:

  • basic - mainly related to some basic fundamental problems which should be mastered smoothly.

  • easy - used for warming up at the very beginning as a fresh man to C language or after long-time un-touching it, which works as a hello-world program.

  • hard - containing something funny and challenging. All the problems inside will be tricky or have quite many various different solutions which also result in difficulty to be innovative enough.


There are many cooler solutions out there. So if you've got one, please do not be any hesitated to share with me. That will be always welcome and funny.


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