1005. Maximize Sum Of Array After K Negations

题目

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

原题链接

解法一

class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
       java.util.Arrays.sort(A);
        int lenA = A.length;
        int lenB=0,lenC=0;
        int result;
        if(A[0]>=0){
            result = arraySum(A,1,lenA-1);
            if(K%2==0)
               result+=A[0];
            else
               result-=A[0];
        }else if(A[lenA-1]<=0){
            result = arraySum(A,K,lenA-1)-arraySum(A,0,1);
        }else{
            result=largestSumInBothPlusMinus(A,K);
        }
       
        return result;
    }
    public int largestSumInBothPlusMinus(int[] A, int K) {
        int result=0;
        int lenA = A.length; 
        int start = 0;
        int end = A.length-1;
        int mid=(start+end)/2;
        int maxLessIndex=-1;
        while(mid>0&&mid+1<lenA){
            if(A[mid]<0){
                if(A[mid+1]<0){
                    start=mid+1;
                    mid=(mid+1+end)/2;
                    if(mid==lenA-2){
                        maxLessIndex=mid;
                        break;
                    }

                }else{
                    maxLessIndex=mid;
                    break;
                }
            }
            else{
               if(A[mid-1]>=0){
                    end=mid-1;
                    mid=(start+mid-1)/2;
                    if(mid==0){
                        maxLessIndex=mid;
                        break;
                    } 
                }else{
                    maxLessIndex=mid-1;
                    break;
                }
            }
        }
        if(K<maxLessIndex+1){
            result=-arraySum(A,0,K-1)+arraySum(A,K,maxLessIndex)+arraySum(A,maxLessIndex+1,lenA-1);
        }else if(K==maxLessIndex+1){
            result=-arraySum(A,0,maxLessIndex)+arraySum(A,maxLessIndex+1,lenA-1);
        }else{
            result=-arraySum(A,0,maxLessIndex)+arraySum(A,maxLessIndex+2,lenA-1);
            int remainK =  K-maxLessIndex-1;
            if(remainK%2==0){
                result+=A[maxLessIndex+1];
            }
            else{
                if(-A[maxLessIndex]<A[maxLessIndex+1]){
                    result=result+A[maxLessIndex+1]+2*A[maxLessIndex];
                }else{
                    result-=A[maxLessIndex+1];
                }
            }
        }
        return result;
        
    }
    public int arraySum(int [] A, int start, int end){
        int sum = 0;
        for(int i=start;i<=end;i++){
            sum+=A[i];
        }
        return sum;
    }
}

解法二：简化求最大的小于0的值得坐标

class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
       java.util.Arrays.sort(A);
        int lenA = A.length;
        int lenB=0,lenC=0;
        int result;
        if(A[0]>=0){
            result = arraySum(A,1,lenA-1);
            if(K%2==0)
               result+=A[0];
            else
               result-=A[0];
        }else if(A[lenA-1]<=0){
            result = arraySum(A,K,lenA-1)-arraySum(A,0,1);
        }else{
            result=largestSumInBothPlusMinus(A,K);
        }
       
        return result;
    }
    public int largestSumInBothPlusMinus(int[] A, int K) {
        int maxLessIndex=-1;
        int result=0;
        int lenA=A.length;
        int i;
        for(i=0;i<lenA;i++){
            if(A[i]<0)
                continue;
            else
                break;
        }
        maxLessIndex=i-1;
        if(K<maxLessIndex+1){
            result=-arraySum(A,0,K-1)+arraySum(A,K,maxLessIndex)+arraySum(A,maxLessIndex+1,lenA-1);
        }else if(K==maxLessIndex+1){
            result=-arraySum(A,0,maxLessIndex)+arraySum(A,maxLessIndex+1,lenA-1);
        }else{
            result=-arraySum(A,0,maxLessIndex)+arraySum(A,maxLessIndex+2,lenA-1);
            int remainK =  K-maxLessIndex-1;
            if(remainK%2==0){
                result+=A[maxLessIndex+1];
            }
            else{
                if(-A[maxLessIndex]<A[maxLessIndex+1]){
                    result=result+A[maxLessIndex+1]+2*A[maxLessIndex];
                }else{
                    result-=A[maxLessIndex+1];
                }
            }
        }
        return result;
        
    }
    public int arraySum(int [] A, int start, int end){
        int sum = 0;
        for(int i=start;i<=end;i++){
            sum+=A[i];
        }
        return sum;
    }
}

解法三：

class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
       java.util.Arrays.sort(A);
       int count=0;
       int sum=0;
       int lenA=A.length;
       int minNum=A[lenA-1];
       for(int i=0;i<lenA;i++){
           if(A[i]<0 && ++count<=K){
               A[i]=-A[i];
           }
           minNum=Math.min(A[i],minNum);
           sum+=A[i];
       }
       if(count>K || (K-count)%2==0){
           return sum;
       }else{
           return sum-2*minNum;
       }

    }
}

注意：if(A[i]<0 && ++count<=K)这里别写成了++count