第一题是k次取反后求数组最大和https://leetcode.cn/problems/maximize-sum-of-array-after-k-negations/description/，使用了两次贪心策略：第一次是优先将绝对值大的负数进行取反，若负数取完后，取反次数仍有剩余，则将小正数进行取反。

class Solution {
static bool cmp(int a, int b){return abs(a) > abs(b);
}
public:int largestSumAfterKNegations(vector<int>& nums, int k) {sort(nums.begin(), nums.end(), cmp);for (int i = 0; i < nums.size(); i++){if (nums[i] < 0 && k > 0) {nums[i] *= -1;k--;}}if (k % 2 == 1) nums[nums.size() - 1] *= -1;int result = 0;for (int i = 0; i < nums.size(); i++){result += nums[i];}return result;}
};

第二题是加油站https://leetcode.cn/problems/gas-station/submissions/502393587/，容易想到的思路是从各个位置开始遍历，知道找出满足要求的起始位置为止，但是代码不好写。

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {for (int i =0; i < cost.size(); i++){int rest = gas[i] - cost[i];int index = (i + 1) % cost.size();while(rest > 0 && index != i){//环形遍历用whilerest += gas[index] - cost[index];index = (index + 1) % cost.size();}if (index == i && rest >= 0) return i;}return -1;}
};

贪心的解法是求出每一站的剩余油量并相加，如果当前遍历发现剩余油量小于0，则重新从下一站开始遍历。

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {int totalsum = 0;for (int i = 0; i < gas.size(); i++){totalsum += gas[i] - cost[i];}if (totalsum < 0) return -1;int cursum = 0;int start = 0;for (int i = 0; i < gas.size(); i++){cursum += gas[i] - cost[i];if (cursum < 0) {start = i + 1; cursum = 0;}}return start;}
};

第三题是分发糖果https://leetcode.cn/problems/candy/description/，需要两次遍历确定当前元素与两边元素的大小关系。从前向后比较当前元素与左侧元素的大小关系，再从后向前比较当前元素与右边元素的大小关系。

class Solution {
public:int candy(vector<int>& ratings) {vector<int> candyVec(ratings.size(), 1);for (int i = 1; i < ratings.size(); i++){if (ratings[i] > ratings[i - 1]) candyVec[i] = candyVec[i - 1] + 1;}for (int i = ratings.size() - 2; i >= 0; i--){if (ratings[i] > ratings[i + 1]){candyVec[i] = max(candyVec[i], candyVec[i + 1] + 1);}}int result = 0;for (int i = 0; i < candyVec.size(); i++){result += candyVec[i];}return result;}
};