Problem:1658. 将 x 减到 0 的最小操作数

思路：逆向思维+滑动窗口

Accode:

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int len = nums.size();
        deque<int> deq;
        long long tot = 0;
        long long sum = 0;
        int ans = 0x3f3f3f3f;
        for(auto i:nums){
            tot+=i;
        }
        for(int i=0;i<len;i++){
            deq.push_back(nums[i]);
            sum+=nums[i];
            while(deq.size() && tot-sum<x){
                sum-=deq.front();
                deq.pop_front();
            }
            if(tot-sum==x) ans = min(ans,(int)(len-deq.size()));
        }
        return ans==0x3f3f3f3f?-1:ans;
    }
};

时间复杂度：$o(n)$