LeetCode Minimum Size Subarray Sum
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题目
思路
O(n)的思路比较简单,直接用两个下标扫一遍即可;
O(nLogn)有点难,个人感觉应该是先得到Sum[i](前i+1)个数的和,因为数字都是正数,那么Sum数组可以用二分查找。我们扫一遍Sum,再二分查找符合条件的前一个Sum的位置即可。
代码
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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