签到题

class Solution {
public:
    int theMaximumAchievableX(int num, int t) {
        return num+t*2;
    }
};

动态规划: 定义$p_i$为跳至$i$处所需的最大跳跃次数, 有状态转移方程 p i = m a x { p j + 1    ∣    0 ≤ j < i , a b s ( n u m s [ i ] − n u m s [ j ] ) ≤ t a r g e t } p_i=max\{ p_j+1 \;|\; 0\le j< i, abs(nums[i]-nums[j])\le target \} pi​=max{pj​+1∣0≤j<i,abs(nums[i]−nums[j])≤target}

class Solution {
public:
    int maximumJumps(vector<int> &nums, int target) {
        int n = nums.size();
        vector<int> p(n, INT32_MIN);
        p[0] = 0;
        for (int i = 1; i < n; i++)
            for (int j = 0; j < i; j++)
                if (p[j] != INT32_MIN && abs(nums[i] - nums[j]) <= target)
                    p[i] = max(p[i], p[j] + 1);
        return p[n - 1] == INT32_MIN ? -1 : p[n - 1];
    }
};

动态规划: 定义$p_{i,0}$为下标$i$选$nums1[i]$情况下$nums3[0,i]$中以$nums3[i]$结尾的最长非递减子数组的长度, 类似地定义
定义$p_{i,1}$为下标$i$选$nums2[i]$情况下$nums3[0,i]$中以$nums3[i]$结尾的最长非递减子数组的长度, $p_{i,j}$只有两种可能的前驱状态$p_{i-1,0}$、$p_{i-1,1}$.

class Solution {
public:
    int maxNonDecreasingLength(vector<int> &nums1, vector<int> &nums2) {
        int n = nums1.size();
        int p[n][2];
        p[0][0] = 1;
        p[0][1] = 1;
        int res = 1;
        for (int i = 1; i < n; i++) {
            p[i][0] = max(nums1[i] >= nums1[i - 1] ? p[i - 1][0] + 1 : 1, nums1[i] >= nums2[i - 1] ? p[i - 1][1] + 1 : 1);
            p[i][1] = max(nums2[i] >= nums1[i - 1] ? p[i - 1][0] + 1 : 1, nums2[i] >= nums2[i - 1] ? p[i - 1][1] + 1 : 1);
            res = max(res, p[i][0]);
            res = max(res, p[i][1]);
        }
        return res;
    }
};

差分数组: 遍历数组同时用差分数组维护当前位置已经减掉的值$cur$, 若遍历到下标为$i$时: 1) $nums[i]-cur>0$ 代表需要在$[i,i+k-1]$这段区间减$cur-nums[i]$(若$i+k-1$越界返回false)，2) $nums[i]-cur<0$ 返回$false$. 遍历结束返回$true$.

class Solution {
public:
    bool checkArray(vector<int> &nums, int k) {
        int n = nums.size();
        vector<int> d(n + 2);
        for (int i = 0, cur = 0; i < n; i++) {
            cur += d[i];//差分数组上的前缀和
            if (nums[i] - cur > 0) {
                if (i + k - 1 >= n)
                    return false;
                d[i + k] -= nums[i] - cur;//更新差分数组d[i+k]
                cur += nums[i] - cur;
            } else if (nums[i] - cur < 0)
                return false;
        }
        return true;
    }
};