小顶堆的使用+数字模拟

1005. Maximize Sum Of Array After K Negations

Easy

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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].

class Solution {
public:
    int largestSumAfterKNegations(vector<int>& A, int K) {
        priority_queue<int,vector<int>,greater<int>> q;
        for(int i = 0;i < A.size();i++){
            q.push(A[i]);
        }
        int num = K;int Result = 0;
        while(num > 0){
            int tmp = q.top();
            q.pop();
            tmp = - tmp;
            q.push(tmp);
            num--;
        }
        while(!q.empty()){
            Result += q.top();
            q.pop();
        }
        return Result;
    }
};

结题思路：无论正数还是负数，取相反数时永远是最小的那个。一个负数越小，取反后相加和就越大。一个正数越小，取反后和的损失越小。就是这样！