C solution, both O(n) & O(nLogn)


  • 4
    N

    O(n):

    int minSubArrayLen(int s, int * nums, int numsSize) {
    	int sum = nums[0], head = 0, tail = 0, minL = numsSize + 1;
    	while (tail < numsSize) {
    		if (tail - head + 1 < minL && sum >= s) minL = tail - head + 1;
    		if (sum >= s) sum -= nums[head++];
    		else sum += nums[++tail];
    		if (head > tail) tail = head;
    	}
    	return minL == numsSize + 1 ? 0 : minL;
    }
    

    O(nLogn):

    // 在不下降的序列中寻找恰好比target小的数出现位置,也即最后一个比target小的数出现的位置
    // search a number that is exactly less than 'target', which means the last number less than 'target'
    int binarySearchIncreaseLastSmaller(int l, int r, int target, int * nums) {  
    	if (l >= r) return -1;
    	while (l < r - 1) {
    		int m = l + ((r - l) >> 1);
    		if (nums[m] < target) l = m;
    		else r = m - 1;
    	}
    	if (nums[r] < target) return r;
    	else if (nums[l] < target) return l;
    	else return -1;
    }
    
    int minSubArrayLen(int s, int * nums, int numsSize) {
    	int * Sum = (int*)malloc(sizeof(int) * (numsSize + 1)), minL = numsSize + 1;
    	Sum[0] = 0;
    	for (int i = 1; i <= numsSize; i++) Sum[i] = Sum[i - 1] + nums[i - 1];
    	for (int i = 1; i <= numsSize; i++) {
    		if (Sum[i] >= s) {
    			int k = Sum[i];
    			int BeforePos = binarySearchIncreaseLastSmaller(0, i, Sum[i] - s + 1, Sum);
    			if (BeforePos != -1 && i - BeforePos < minL) minL = i - BeforePos;
    		}
    	}
    	return minL == numsSize + 1 ? 0 : minL;
    }

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