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

解題思路

要讓總和最大，需要每次都取反最小的元素。一個優(yōu)化點是：如果最小的元素是正數(shù)，可以直接得到最終結(jié)果，無需每次都遍歷找到最小元素（最小正數(shù)取反之后還是最小的）。

實現(xiàn)代碼

實現(xiàn)1：

//Runtime: 2 ms, faster than 92.17% of Java online submissions for Maximize Sum Of Array After K Negations.
//Memory Usage: 38 MB, less than 100.00% of Java online submissions for Maximize Sum Of Array After K Negations.
class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
        int sum = getSum(A);
        while (K-- > 0) {
            int index = getIndexOfSmallest(A);
            A[index] = -A[index];
            sum += 2 * A[index];
        }
        return sum;
    }

    public int getSum(int[] nums) {
        int result = 0;
        for (int num : nums) {
            result += num;
        }
        return result;
    }

    public int getIndexOfSmallest(int[] nums) {
        int smallest = Integer.MAX_VALUE;
        int index = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < smallest) {
                smallest = nums[i];
                index = i;
            }
        }
        return index;
    }
}

實現(xiàn)2：

//Runtime: 1 ms, faster than 99.93% of Java online submissions for Maximize Sum Of Array After K Negations.
//Memory Usage: 38 MB, less than 100.00% of Java online submissions for Maximize Sum Of Array After K Negations.
class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
        int sum = getSum(A);
        while (K > 0) {
            int index = getIndexOfSmallest(A);
            if (A[index] < 0) {
                K--;
                A[index] = -A[index];
                sum += 2 * A[index];
            } else {
                if (K % 2 == 0) {
                    return sum;
                } else {
                    return sum - 2 * A[index];
                }
            }
        }
        return sum;
    }

    public int getSum(int[] nums) {
        int result = 0;
        for (int num : nums) {
            result += num;
        }
        return result;
    }

    public int getIndexOfSmallest(int[] nums) {
        int smallest = Integer.MAX_VALUE;
        int index = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < smallest) {
                smallest = nums[i];
                index = i;
            }
        }
        return index;
    }
}