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目录

题目描述

方法一、使用小根堆

方法二、排序

方法三、哈希

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题目描述

1005. Maximize Sum Of Array After K Negations

三种方法都是贪心。 

方法一、使用小根堆

class Solution {
public:int largestSumAfterKNegations(vector<int>& nums, int k) {int sum = std::accumulate(nums.begin(),nums.end(),0);priority_queue<int,vector<int>,std::greater<int>> minHeap(nums.begin(),nums.end());while(k--){int min = minHeap.top();minHeap.pop();minHeap.push(-min);sum -= (2*min);}return sum;}
};

还可以优化一下。处理完负数后，当剩余数字全部是非负数的时候，可以一次处理结束。

class Solution {
public:int largestSumAfterKNegations(vector<int>& nums, int k) {int sum = std::accumulate(nums.begin(),nums.end(),0);priority_queue<int,vector<int>,std::greater<int>> minHeap(nums.begin(),nums.end());while(k){int min = minHeap.top();if(min>=0){if(k%2==1)sum -=(2*min);break;}minHeap.pop();minHeap.push(-min);sum -= (2*min);k--;}return sum;}
};

方法二、排序

按照绝对值从大到小排序。

class Solution {struct Cmp{bool operator()(int a,int b) const{return abs(a)>abs(b);}};
public:int largestSumAfterKNegations(vector<int>& nums, int k) {sort(nums.begin(),nums.end(),Cmp());int n = nums.size();for(int i = 0;i < n;i++){if(k>0 && nums[i] < 0){nums[i]*=-1;k--;}if(k ==0)break;}if(k%2==1) nums[n-1]*=-1;return accumulate(nums.begin(),nums.end(),0);}
};

方法三、哈希

由于-100 <= nums[i] <= 100，故方法二的排序可以用哈希替代。

class Solution {
public:int largestSumAfterKNegations(vector<int>& nums, int k) {int sum = 0;unordered_map<int,int> count;for(int num:nums){count[num]++;sum += num;}for(int i = -100;i <0;i++){if(count[i] > 0){int operation_cnt = min(count[i],k);sum -= operation_cnt*2*i;k -= operation_cnt;count[i] -= operation_cnt;count[-i]+= operation_cnt;if(k == 0)break;}}if(k>0 && k%2==1 && count[0]==0){for(int i = 0;i <=100;i++){if(count[i]>0){sum -= 2*i;break;}}}return sum;}
};