[LeetCode] Jump Game II

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

f[i,j]表示 从i 到j 的步骤，

f[i,j] = min{ f[i,k] + 1, if(A[k] > j-k)}

我们可以用f[j] 表示上述的f[0,j ] ，所以状态转移为：f[j] = min{ f[k] + 1, if(A[k] > j-k)}

 

第一次的版本，居然 Status:Runtime Error，后来查看，由于内存泄露引起的，f申请的堆内存没有释放。

class Solution {
    public:
        int jump(int A[], int n)
        {   
            int* f = new int(n);
            f[0] = 0;
            for(int i = 1; i < n; i++)
                f[i] = INT_MAX;
    
            for(int i = 1; i < n; i++)
            {   
                for(int j = 0; j < i; j++)
                {   
                    if(A[j] >= (i-j))// can jump from j to i 
                        f[i] = min(f[i], f[j] + 1); 
                }   
            }   
            //printArray(f, n); 
            return f[n-1];

        }   
};

 

第二版本：用vecotr 代替 堆申请内存， 但是 Submission Result: Time Limit Exceeded 

Submission Result: Time Limit ExceededMore Details 

class Solution {
    public:
        int jump(int A[], int n)
        {
            vector<int> f;
            f.resize(n);
            f[0] = 0;
            for(int i = 1; i < n; i++)
                f[i] = INT_MAX;

            for(int i = 1; i < n; i++)
            {
                for(int j = 0; j < i; j++)
                {
                    if(A[j] >= (i-j))// can jump from j to i 
                        f[i] = min(f[i], f[j] + 1);
                }
            }
            return f[n-1];

        }
};

 方法3，贪心法  copy from https://github.com/haoel/leetcode/blob/master/src/jumpGame/jumpGame.II.cpp

比如第一步能走2， 那么久从A[1] +1 和 A[2] + 2 中选取最大的，这样持续下去，

不如array A = [2,3,1,1,4]，A[1]+1 = 4， A[2]+2 = 3 所以选择1，因为从A[1]出发覆盖的范围更大，下次从A[1]出发，A[1]覆盖的范围是3， 那么就是A[2]+2，A[3]+3，A[4]+4 中选大的，迭代下去。。

class Solution {
    public:
//Acutally, using the Greedy algorithm can have the answer
int jump(int A[], int n) {
    if (n<=1) return 0;
    
    int steps = 0;
    int coverPos = 0;
    for (int i=0; i<=coverPos&& i<n; ){
        if (A[i]==0) return -1; 
        if(coverPos < A[i]+i){
            coverPos = A[i]+i;
            steps++;
        }
        if (coverPos >= n-1){
            return steps;
        }
        //Greedy: find the next place which can have biggest distance
        int nextPos=0;
        int maxDistance=0;
        for(int j=i+1; j<=coverPos; j++){
            if ( A[j]+j > maxDistance ) {
                maxDistance = A[j]+j;
                nextPos = j;
            }
        }
        i = nextPos;
    }
    return steps;
}
};

 贪心2，好理解点的算法

class Solution {
    public:
        int jump(int A[], int n) {
            
            if(n == 1) return 0; //只有一个元素，只需啊0步骤
            int maxReach = 0;// the most rigth we can reach

            int steps = 0;

            for(int i = 0; i <= maxReach ; /* i++ */)//i means current position
            {
                steps ++;
                int newMaxReach = 0;
                for(int j = i; j <= maxReach && j < n; j++ )// count i
                {
                    if(A[j] + j > newMaxReach)
                    {
                        newMaxReach = A[j] + j;
                        i = j + 1; // begin with j+1
                    }
                }

                if(newMaxReach >= n-1)
                {
                    return steps;
                }
                else if(newMaxReach > maxReach)
                {
                    // update the maxReach
                    maxReach = newMaxReach;
                }
            }

            return 0;

        }
};