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