今日任务

122.买卖股票的最佳时机II

• 题目链接： https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/description/
• 题目描述：
  

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        // int profit = 0;
        // strong_ordering cmp_less = strong_ordering::less;
        // strong_ordering cmp_greater = strong_ordering::greater;
        // int buy = 0;
        // bool is_buy = false;
        // for(int i = 1; i < prices.size(); i++){
        //     if(!is_buy && prices[i - 1] <=> prices[i] == cmp_less){
        //         buy = prices[i - 1];
        //         is_buy = true;
        //     }else if(is_buy && prices[i - 1] <=> prices[i] == cmp_greater){
        //         profit += prices[i - 1] - buy;
        //         is_buy = false;
        //     }
        // }
        // if(is_buy && prices.back() > buy){
        //     profit += prices.back() - buy;
        // }
        // return profit;

        int ans = 0;
        int pre = 0;
        int n = prices.size();
        bool flag = false;
        for(int i = 1; i < n; i++){
            int a = prices[i - 1], b = prices[i];
            if(!flag && a < b){
                pre = a;
                flag = !flag;
            }else if(flag && a > b){
                ans += a - pre;
                flag = !flag;
            }
        }
        if(flag && prices.back() > pre){
            ans += prices.back() - pre;
        }
        return ans;
    }
};

55. 跳跃游戏

• 题目链接： https://leetcode.cn/problems/jump-game/description/
• 题目描述：
  

Code

class Solution {
public:
    bool canJump(vector<int>& nums) {
        // int target = 1;
        // for(auto it = nums.rbegin() + 1; it != nums.rend(); it++){
        //     if(*it >= target){
        //         target = 1;
        //     }else{
        //         target++;
        //     }
        // }
        // return target == 1;

        // int mx_idx = 0;
        // for(int i = 0; i < nums.size(); i++){
        //     if(i > mx_idx){
        //         return false;
        //     }else{
        //         mx_idx = max(mx_idx, i + nums[i]);
        //     }
        // }
        // return true;

        int mx_idx = 0;
        for(int i = 0; mx_idx < nums.size() - 1; i++){
            if(i > mx_idx){
                return false;
            }else{
                mx_idx = max(mx_idx, i + nums[i]);
            }
        }
        return true;
    }
};

45.跳跃游戏II

• 题目链接： https://leetcode.cn/problems/jump-game-ii/description/
• 题目描述：
  

Code

class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size();
        int mx_index = nums[0];
        int ans = 0;
        for(int i = 1; i < n; ){
            int mx = 0;
            while(i < n && i <= mx_index){
                mx = max(mx, i + nums[i]);
                i++;
            }
            mx_index = mx;
            ans++;
        }
        return ans;
    }
};

1005. K 次取反后最大化的数组和

• 题目链接： https://leetcode.cn/problems/maximize-sum-of-array-after-k-negations/
• 题目描述：
  

Code

class Solution {
public:
    int largestSumAfterKNegations(vector<int>& nums, int k) {
        priority_queue<int, vector<int>, greater<int>> que(nums.begin(), nums.end());
        int sum = reduce(nums.begin(), nums.end(), 0);
        while(k--){
            int t = que.top();
            que.pop();
            if(t < 0){
                que.push(-t);
                sum += -2 * t;
            }
            else if(t > 0){
                sum += (k + 1) % 2 == 0 ? 0 : -t * 2;
                break;
            }
            else {
                break;
            }
        }

        return sum;
    }
};