3086. 拾起 K 个 1 需要的最少行动次数

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题目链接：3086. 拾起 K 个 1 需要的最少行动次数

代码如下：

//参考链接:https://leetcode.cn/problems/minimum-moves-to-pick-k-ones/solutions/2692009/zhong-wei-shu-tan-xin-by-endlesscheng-h972
class Solution
{
public:long long minimumMoves(vector<int>& nums, int k, int maxChanges){vector<int> pos;int c = 0;//nums中连续1的长度for (int i = 0; i < nums.size(); i++){if (nums[i] == 0) { continue; }pos.push_back(i);//记录1的位置c = max(c, 1);if (i > 0 && nums[i - 1] == 1){if (i > 1 && nums[i - 2] == 1){c = 3;//有3个连续的1}else{c = max(c, 2);//有2个连续的1}}}c = min(c, k);if (maxChanges >= k - c){//其余k-c个1可以全部用两次操作得到return max(c - 1, 0) + (k - c) * 2;}int n = pos.size();vector<long long> sum(n + 1);for (int i = 0; i < n; i++){sum[i + 1] = sum[i] + pos[i];}long long res = LLONG_MAX;//除了maxChanges个数可以用两次操作得到，其余的1只能一步步移动到pos[iint size = k - maxChanges;for (int right = size; right <= n; right++){//s1+s2是j在[left,right)中所有pos[j]到index=pos[(left+right)/2] 的距离之和int left = right - size;int i = left + size / 2;long long index = pos[i];long long s1 = index * (i - left) - (sum[i] - sum[left]);long long s2 = sum[right] - sum[i] - index * (right - i);res = min(res, s1 + s2);}return res + maxChanges * 2;}
};