(贪心 优先队列) leetcode1005. Maximize Sum Of Array After K Negations

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].

Input: A = [3,-1,0,2], K = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].

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. 1 <= A.length <= 10000
2. 1 <= K <= 10000
3. -100 <= A[i] <= 100

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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;
    }
};

(贪心 优先队列) leetcode1005. Maximize Sum Of Array After K Negations

原文：https://www.cnblogs.com/Weixu-Liu/p/10822501.html