LeetCode题解:Jump Game

Jump Game


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.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.


思路:

直接考虑的方式是这样的,最后一个元素标记为true,然后依次考察之前的元素,每个之前的元素如果在前进范围内有标记为true的元素,则标记为true,否则标记为false。这样直到第一个元素,返回第一个元素的值。

题解:

class Solution {
public:
    bool canJump(int A[], int n) {
        // store the maximum index can be reached from the i-th index
        vector<bool> reachable(n);
        reachable[n-1] = true;  // it can always reach itself
        
        auto reach_forward=[&reachable,n](const int index, const int dist) -> bool {
            for(int fwd = index+1; fwd <= min(dist, n-1); ++fwd)
                if (reachable[fwd]) 
                    return true;
            return false;
        };
        
        // reverse iterate every previous element 
        for(int ri = n-2; ri>=0; --ri)
            reachable[ri]=reach_forward(ri, A[ri]);
            
        return reachable[0];
    }
};
思路:

上述题解能够给出正确的解,但是时间复杂度为O(n^2)。如果有O(n)的解法更好。

从正向考察每个元素,记能到达的最远元素为so_far。如果so_far>=n-1,说明可以到达最终元素,如果已经考察到记为so_far的元素,仍不能令so_far >= n-1,说明无法到达最终元素。

题解:

class Solution {
public:
    bool canJump(int A[], int n) {
        // store the farthest element can be reached
        int so_far = 0;
        
        for(int iter=0; iter<=so_far; ++iter)
            if ( (so_far = max(so_far, A[iter] + iter)) >= n-1 )
                return true;
        return false;
    }
};


你可能感兴趣的:(LeetCode)