leetcode打卡#day36 1005. K 次取反后最大化的数组和、134. 加油站、135. 分发糖果

1005. K 次取反后最大化的数组和

class Solution {
private:
    //将数组元素按绝对值从大到小排列
    static bool cmp(int a, int b) {
        return abs(a) > abs(b);
    }
public:
   int largestSumAfterKNegations(vector<int>& nums, int k) {
        //对数据进行排序
        sort(nums.begin(), nums.end(), cmp);
        int sum = 0;
        for (int i = 0; i<nums.size(); i++) {
            if(nums[i] < 0 && k > 0){
                nums[i] = -nums[i];
                k--;
            } 
        }
        //剩下的加上
        if (k%2 == 1) {
            nums[nums.size()-1] *= -1;
        }
        for (int it: nums) {
            sum += it;
        }
        return sum;

    }
};

134. 加油站

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int curSum = 0;
        //走完一圈之后还剩下多少
        int remain = 0;
        //开始的地方
        int start = 0;
        for (int i = 0; i < gas.size(); i++) {
            int cha = gas[i] - cost[i];
            curSum += cha;
            remain += cha;
            if (curSum < 0) {
                start = i+1; //选择新的起始点
                curSum = 0;//curSum清空
            }
        }
        //当油量 < 消耗时, 没有足够汽油走完全程
        if (remain < 0) return -1;

        return start;
    }
};

135. 分发糖果

class Solution {
public:
    int candy(vector<int>& ratings) {
        int n = ratings.size();
        int total = 0;
        vector<int> candyVec(n, 1);
        //从下标为1开始判断, 从左往右
        for (int i = 1; i < n; i++) {
            if (ratings[i] > ratings[i-1]) candyVec[i] = candyVec[i-1] + 1;
        }
        //从右往左
        for (int i = n-2; i >= 0; i--) {
            if (ratings[i] > ratings[i+1]) candyVec[i] = max(candyVec[i], candyVec[i+1]+1);
        }
        //总和所需要的糖果数
        for (int it: candyVec) {
            total += it;
        }
        return total;
    }
};