Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We may choose the same index i multiple times.)
Return the largest possible sum of the array after modifying it in this way.
Example 1:
Input: A = [4,2,3], K = 1 Output: 5 Explanation: Choose indices (1,) and A becomes [4,-2,3].Example 2:
Input: A = [3,-1,0,2], K = 3 Output: 6 Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].Example 3:
Input: A = [2,-3,-1,5,-4], K = 2 Output: 13 Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].
Note:
1 <= A.length <= 100001 <= K <= 10000-100 <= A[i] <= 100-------------------------------------------------------------------------------------------------------------------
这个题是贪心题，不过，要注意在有负数和整数的情况下，多次将其中的最小的数（负数）反转为正数时，最后得到一系列整数时，还需要进行排序。emmm，这个地方不是可以用优先队列吧，所以用了优先队列，不过在时间复杂度上比较吃亏了。。。
C++代码：
class Solution { public: int largestSumAfterKNegations(vector<int>& A, int K) { priority_queue<int,vector<int>,greater<int> > pq; for(int num:A){ pq.push(num); } int ans = 0; while(ans < K){ int a = pq.top();pq.pop(); a = -a; pq.push(a); ans++; } int sum = 0; while(!pq.empty()){ sum += pq.top(); pq.pop(); } return sum; } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10822501.html
