将序列复制一遍接到序列后(展开成2n), 问题转化为在2n长度的序列中寻找和为x的最小长度子序列。注意边界条件的判定(i >= n -1 && j <= n)
class Solution {
public:
int minOperations(vector<int>& nums, int x) {
int n = nums.size();
vector<int> arr(2*n);
for (int i = 0; i < n; i++)
arr[i] = arr[i+n] = nums[i];
int ans = INT_MAX, sum = 0;
for (int i = 0, j = 0; i<2*n; i++){
sum += arr[i];
while (j < i && sum > x)
sum -= arr[j++];
if (i < n-1 || j > n) continue;
if (sum == x)
ans = min(ans, i - j + 1);
}
return ans == INT_MAX || ans > n ? -1 : ans;
}
};