day28 贪心2

122.买卖股票的最佳时机II

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        int buyPrice = prices[0];
        for (int i = 1; i < prices.size(); i ++) {
            if (prices[i] > buyPrice) {
                res += prices[i] - buyPrice;
            }
            buyPrice = prices[i];
        }
        return res;
    }
};

55. 跳跃游戏

class Solution {
public:
    bool canJump(vector<int>& nums) {
        int maxSteps = nums[0];
        for (int i = 1; i < nums.size(); i ++) {
            if (maxSteps < i) {
                return false;
            }
            maxSteps = max(maxSteps, i + nums[i]);
        }
        return true;
    }
};

45.跳跃游戏II

class Solution {
public:
    int jump(vector<int>& nums) {
        if (nums.size() == 1) {
            return 0;
        }
        int res = 1;
        int cover = nums[0];
        int i = 1;
        while (i <= cover) {
            if (cover >= nums.size() - 1) {
                break;
            }
            int maxIndex = i;
            int maxSteps = i + nums[i];
            for (int j = i; j <= cover; j ++) {
                if (j + nums[j] >= maxSteps) {
                    maxSteps = j + nums[j];
                    maxIndex = j;
                }
            }
            i = maxIndex;
            cover = maxSteps;
            res ++;
        }
        return res;
    }
};

1005.K次取反后最大化的数组和

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