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Update 1005.K次取反后最大化的数组和.md
与C++相同的思路,不过排序的耗时更长
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@ -99,6 +99,33 @@ public:
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Java:
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Java:
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```java
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class Solution {
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public int largestSumAfterKNegations(int[] nums, int K) {
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// 将数组按照绝对值大小从大到小排序,注意要按照绝对值的大小
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nums = IntStream.of(nums)
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.boxed()
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.sorted((o1, o2) -> Math.abs(o2) - Math.abs(o1))
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.mapToInt(Integer::intValue).toArray();
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int len = nums.length;
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for (int i = 0; i < len; i++) {
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//从前向后遍历,遇到负数将其变为正数,同时K--
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if (nums[i] < 0 && k > 0) {
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nums[i] = -nums[i];
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k--;
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}
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}
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// 如果K还大于0,那么反复转变数值最小的元素,将K用完
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if (k % 2 == 1) nums[len - 1] = -nums[len - 1];
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int result = 0;
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for (int a : nums) {
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result += a;
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}
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return result;
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}
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}
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```
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```java
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```java
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class Solution {
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class Solution {
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public int largestSumAfterKNegations(int[] A, int K) {
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public int largestSumAfterKNegations(int[] A, int K) {
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