力扣 1005.K次取反后最大化的数组和

class Solution {public int largestSumAfterKNegations(int[] nums, int k) {// 将基本类型的int数组转换成IntStream，以便进行流操作。nums = Arrays.stream(nums)// 将IntStream中的int元素转换（装箱）为Integer对象，这是必须的因为后面的sorted方法需要对象来进行比较。.boxed()// 对流中的元素进行排序。使用自定义的比较器，该比较器基于元素的绝对值进行降序排序。// o1和o2是比较的两个元素，Math.abs(o2) - Math.abs(o1)计算结果如果为正数，则o2排在o1前面，实现降序。.sorted((o1, o2) -> Math.abs(o2) - Math.abs(o1))// 将流中的Integer对象转换回基本类型int，以便可以将流转换回数组。.mapToInt(Integer::intValue)// 将流中的元素收集到一个新的int数组中。.toArray();for(int i = 0; i < nums.length && k > 0; i++) {if(nums[i] < 0) {nums[i] = nums[i] * -1;k--;}}if(k == 0 || k % 2 == 0) {return Arrays.stream(nums).sum();}else {nums[nums.length - 1] = nums[nums.length - 1] * -1;return Arrays.stream(nums).sum();}    }
}

力扣 134. 加油站

class Solution {public int canCompleteCircuit(int[] gas, int[] cost) {int[] res = new int[gas.length];for(int i = 0; i < gas.length; i++) {res[i] = gas[i] - cost[i];}if(Arrays.stream(res).sum() < 0) {return -1;}else {int idx = 0;int sum = 0;while(idx < res.length) {int i;for(i = idx; i < res.length; i++) {sum += res[i];if(sum < 0) {sum = 0;idx = i + 1;break;}}if(i == res.length) return idx;}}return -1;}
}

力扣 135. 分发糖果

class Solution {public int candy(int[] ratings) {int[] LToR = new int[ratings.length];LToR[0] = 1;for(int i = 1; i < ratings.length; i++) {if(ratings[i] > ratings[i - 1]) {LToR[i] = LToR[i - 1] + 1;}else {LToR[i] = 1;}}int[] RToL = new int[ratings.length];RToL[ratings.length - 1] = 1;for(int i = ratings.length - 2; i >= 0; i--) {if(ratings[i] > ratings[i + 1]) {RToL[i] = RToL[i + 1] + 1;}else {RToL[i] = 1;}}int[] result = new int[ratings.length];for(int i = 0; i < ratings.length; i++) {result[i] = Math.max(LToR[i], RToL[i]);}int sum = Arrays.stream(result).sum();return sum;}
}