Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
- Example:
- Input: s = 7, nums = [2,3,1,2,4,3]
- Output: 2
- Explanation: the subarray [4,3] has the minimal length under the problem constraint.
- Follow up:
- If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
$O(n)$ 代码:
- class Solution {
- public:
- int minSubArrayLen(int s, vector<int>& nums) {
- if(nums.empty()) return 0;
- int n = nums.size();
- int cnt = n + 1;
- int R = 0, L = 0, sum = 0;
- while(R <n) {
- while(sum < s && R < n)
- sum += nums[R ++];
- while(sum>= s) {
- cnt = min(cnt, R - L);
- sum -= nums[L ++];
- }
- }
- if(cnt == n + 1) return 0;
- return cnt;
- }
- };
- View Code
$O(n^2)$ 代码:
- class Solution {
- public:
- int minSubArrayLen(int s, vector<int>& nums) {
- if(nums.empty()) return 0;
- int n = nums.size();
- int cnt = n + 1;
- for(int i = 0; i < n; i ++) {
- int temp = i;
- int ans = 0;
- while(ans < s && temp < n) {
- ans += nums[temp];
- temp ++;
- }
- if(ans < s) continue;
- else cnt = min(cnt, temp - i);
- }
- if(cnt == n + 1) return 0;
- return cnt;
- }
- };
- View Code
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