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 -1 instead.
Given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint.
Challenge
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
LeetCode上的原题，请参见我之前的博客Minimum Size Subarray Sum。
解法一：
class Solution { public: /** * @param nums: a vector of integers * @param s: an integer * @return: an integer representing the minimum size of subarray */ int minimumSize(vector<int> &nums, int s) { int res = INT_MAX, sum = 0, left = 0; for (int i = 0; i < nums.size(); ++i) { sum += nums[i]; if (sum >= s) { while (left < i && sum >= s) { res = min(res, i - left + 1); sum -= nums[left++]; } } } return res == INT_MAX ? -1 : res; } };
解法二：
class Solution { public: /** * @param nums: a vector of integers * @param s: an integer * @return: an integer representing the minimum size of subarray */ int minimumSize(vector<int> &nums, int s) { int res = INT_MAX, n = nums.size(); vector<int> sums(n + 1, 0); for (int i = 1; i < n + 1; ++i) sums[i] = sums[i - 1] + nums[i - 1]; for (int i = 0; i < n + 1; ++i) { int left = i + 1, right = n, t = sums[i] + s; while (left <= right) { int mid = left + (right - left) / 2; if (sums[mid] < t) left = mid + 1; else right = mid - 1; } if (left == n + 1) break; res = min(res, left - i); } return res == INT_MAX ? -1 : res; } };
本文转自博客园Grandyang的博客，原文链接：最小子数组和的大小[LintCode] Minimum Size Subarray Sum ，如需转载请自行联系原博主。
