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力扣1005.k次取反后最大化的数组和

链接: link

思路

既然要求最大和，那么不妨先给数组排个序，如果有负数，先处理负数从前往后给数组取反，如果负数处理完后k还有次数，此时数组全是正数了，只需要对第一个元素取反即可，无非就是奇数次或者偶数次取反操作。最终求和即可。

方法1:

class Solution {public int largestSumAfterKNegations(int[] nums, int k) {if (nums.length == 1)return nums[0];int ans = 0;Arrays.sort(nums);// 先处理负数for (int i = 0; i < nums.length && k > 0; i++) {if (nums[i] < 0) {nums[i] = -nums[i];k--;}}// 如果k还有次数if (k % 2 == 1) {Arrays.sort(nums);nums[0] = -nums[0];}for (int num : nums) {ans += num;}return ans;}
}

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