算法D33 | 贪心算法3 | 1005.K次取反后最大化的数组和 134. 加油站 135. 分发糖果

本题简单一些，估计大家不用想着贪心 ，用自己直觉也会有思路。 

代码随想录

Python:

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        nums.sort(key=lambda x: abs(x), reverse=True) # 关键：按绝对值降序排列
        for i in range(len(nums)):
            if nums[i]<0 and k>0:
                nums[i] = -nums[i]
                k -= 1
        if k%2==1: 
            nums[-1] *= -1
        return sum(nums)

C++：

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); //注意：这里cmp是static method
        for (int i=0; isize(); i++) {
            if (nums[i]<0 && k>0) {
                nums[i] *= -1;
                k--;
            }
        }
        if (k%2==1) nums[nums.size()-1] *= -1;
        int ans = 0;
        for (int a:nums) ans+=a;
        return ans;
    }
};

134. 加油站 

本题有点难度，不太好想，推荐大家熟悉一下方法二 

代码随想录

Python:

情况三是比较难想到的，从后向前看如何覆盖cum_sum of net gas.

• 情况三：如果累加的最小值是负数，汽车就要从非0节点出发，从后向前，看哪个节点能把这个负数填平，能把这个负数填平的节点就是出发节点。

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        cum_sum = 0
        min_net = float('inf')
        for i in range(len(gas)):
            cum_sum += gas[i] - cost[i]
            if cum_sum < min_net:
                min_net = cum_sum
        if cum_sum < 0: return -1
        if min_net > 0: return 0
        for i in reversed(range(len(gas))):
            min_net += gas[i] - cost[i]
            if min_net >= 0:
                return i
        return -1

C++：

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

135. 分发糖果 

本题涉及到一个思想，就是想处理好一边再处理另一边，不要两边想着一起兼顾，后面还会有题目用到这个思路 

代码随想录

Python:

class Solution:
    def candy(self, ratings: List[int]) -> int:
        n = len(ratings)
        if n == 1: return 1
        ans = [1] * n 
        for i in range(1, n): # 从前往后
            if ratings[i] > ratings[i-1]:
                ans[i] = ans[i-1] + 1
        for i in reversed(range(n-1)): # 从后往前
            if ratings[i] > ratings[i+1]:
                ans[i] = max(ans[i], ans[i+1]+1)
        return sum(ans)

C++：

class Solution {
public:
    int candy(vector<int>& ratings) {
        if (ratings.size()==1) return 1;
        vector<int> candyVec(ratings.size(), 1);    
        for (int i=1; isize(); 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 ans = 0;
        for (int c: candyVec) ans += c;
        return ans;
    }
};