C++前缀和算法的应用：得到连续 K 个 1 的最少相邻交换次数 原理源码测试用例

class Solution {
public:
int minMoves(vector& nums, int k) {
m_c = nums.size();
vector vOneIndex;
for (int i = 0; i < m_c ; i++)
{
if (1 == nums[i])
{
vOneIndex.emplace_back(i);
}
}
vector v0Dis = { 0 };//记录nums[0,i)中,nums[i]等于0时 i之和,也就是将所有nums[i]移到0处
for (int i = 0; i < m_c; i++)
{
long long llAdd = (0 == nums[i]) ? i : 0;
v0Dis.emplace_back(llAdd+v0Dis.back());
}
vector v0Num = { 0 };//记录nums[0,i)中0的个数
for (const auto& n : nums)
{
v0Num.emplace_back(v0Num.back() + (0==n));
}
long long llRet = INT_MAX;
int m = 0;
for (int i = 0; i + k - 1 < vOneIndex.size(); i++)
{
const int left = vOneIndex[i];
const int r = vOneIndex[i + k - 1];
if (m < left)
{
m = left + 1;
}
for (; m < r; m++)
{
if (1 == nums[m])
{
continue;
}
//[left,m)中的0 左移
const int iLeft0Num = v0Num[m] - v0Num[left];
//[m,end)中的0右移，由于nums[end]是1，所以[m,end]中的0，就是[m,end)中的0
const int iRight0Num = v0Num[r] - v0Num[m];
const int iLeftCur = m - left - iLeft0Num;
const int iRightCur = r - m - (iRight0Num - 1);
if (iRightCur <= iLeftCur)
{
break;
}
}
//m 等于r，也符合下面的逻辑[left,r)和[r,r)，右移为空
const long long iLeft0Num = v0Num[m] - v0Num[left];
const long long iRight0Num = v0Num[r] - v0Num[m];
const long long llLeftMove = v0Dis[m] - v0Dis[left] - left * iLeft0Num - (iLeft0Num - 1) * iLeft0Num / 2;
const long long llRightMove = r * iRight0Num - (v0Dis[r] - v0Dis[m]) - (iRight0Num - 1) * iRight0Num / 2;
llRet = min(llRet, llLeftMove + llRightMove);
}
return llRet;
}
int m_c;
};

template
void Assert(const vector& v1, const vector& v2)
{
if (v1.size() != v2.size())
{
assert(false);
return;
}
for (int i = 0; i < v1.size(); i++)
{
assert(v1[i] == v2[i]);
}
}
template
void Assert(const T& t1, const T& t2)
{
assert(t1 == t2);
}
int main()
{
vector nums = { 1, 0, 0, 1, 0, 1 };
int k = 2;
auto res = Solution().minMoves(nums,k);
Assert(1, res);
nums = { 1, 0, 0, 0, 0, 0, 1, 1 };
k = 3;
res = Solution().minMoves(nums, k);
Assert(5, res);
nums = { 1,1,0,1 };
k = 2;
res = Solution().minMoves(nums, k);
Assert(0, res);