leetcode1658 将 x 减到 0 的最小操作数

思路1：

前缀和+后缀和。

实现1：

class Solution
{
public:
    const int INF = 0x3f3f3f3f;
    int minOperations(vector<int>& nums, int x)
    {
        int n = nums.size(), suffix = 0;
        unordered_map<int, int> mp;
        mp[0] = n;
        for (int i = n - 1; i >= 0; i--)
        {
            suffix += nums[i];
            mp[suffix] = i;
        }
        int prefix = 0;
        int res = INF;
        if (mp.count(x)) res = n - mp[x];
        for (int i = 0; i < n; i++)
        {
            mp.erase(suffix); suffix -= nums[i];
            prefix += nums[i];
            if (mp.count(x - prefix))
            {
                res = min(res, i + 1 + n - mp[x - prefix]);
            }
        }
        return res == INF ? -1: res;
    }
};

思路2：

用滑动窗口在数组中找到和等于sum-x的最长的子段（sum是数组所有元素的和）。

实现2：

class Solution
{
public:
    int minOperations(vector<int>& nums, int x)
    {
        int n = nums.size(), sum = 0, slow = 0, res = -1;
        int tot = accumulate(nums.begin(), nums.end(), 0);
        if (tot == x) return n;
        x = tot - x;
        for (int i = 0; i < n; i++)
        {
            sum += nums[i];
            if (sum >= x)
            {
                while (slow <= i and sum >= x)
                {
                    if (sum == x) res = max(res, i - slow + 1);
                    sum -= nums[slow++];
                }
            }
        }
        return res == -1 ? -1: n - res;
    }
};