Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.
For example, given the array [2,3,1,2,4,3]
and s = 7
,
the subarray [4,3]
has the minimal length under the problem constraint.
[balabala] O(N)的做法是两指针法或者叫滑动窗口法, right 向前走直到sum > s, 然后left 向前走直到sum < s, 同时更新minLen. 网上找的O(nlogn) 方法不理解,先备忘着,实现之一如Method3, 实现2参见http://www.tuicool.com/articles/miQVzm3
// Method 2: 伸缩滑动窗口法,O(N), 实现2 public int minSubArrayLen2(int s, int[] nums) { if (nums == null || nums.length == 0) return 0; int left = 0, right = 0, sum = 0, minLen = nums.length; while (right < nums.length) { while (sum < s && right < nums.length) { sum += nums[right++]; } while (sum >= s && left < right) { minLen = Math.min(minLen, right - left); sum -= nums[left++]; } } return left > 0 ? minLen : 0; } // Method 1: 伸缩滑动窗口法,O(N),实现1 public int minSubArrayLen(int s, int[] nums) { if (nums == null || nums.length == 0) return 0; int left = 0, right = 0, sum = 0, minLen = nums.length; while (left <= right) { if (sum < s && right < nums.length) { sum += nums[right++]; } else if (sum < s && right == nums.length) { break; } if (sum >= s) { minLen = Math.min(minLen, right - left); if (minLen == 1) return 1; sum -= nums[left++]; } } return left > 0 ? minLen : 0; } // 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 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; } // Method 3: O (nlogn) 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; }