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];elseresult-=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];elseresult-=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;elsebreak;}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