[LeetCode]Jump Game

最新推荐文章于 2025-05-22 14:52:23 发布

原创最新推荐文章于 2025-05-22 14:52:23 发布 · 1.2k 阅读

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本文探讨了如何通过改进算法来解决跳跃问题，包括使用排序和遍历优化，以达到更高效的时间复杂度。详细解释了两种解决方案：一种是基于段重叠的O(nlogn)方法，另一种是O(n^2)的直接方法，比较了它们的性能并提供了代码实现。

class Solution {
//O(n^2) TLE
//using segment overlap O(nlogn)
public:
	struct Seg
	{
		int x, y;
		Seg(int _x=0, int _y=0):x(_x),y(_y){}
		bool operator < (const Seg& rhs) const
		{
			if(x != rhs.x) return x < rhs.x;
			else return y < rhs.y;
		}
	};
	bool canJump(int A[], int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(0 == n) return false;
		vector<Seg> seg(n);
		for (int i = 0; i < n; ++i)
			seg[i] = Seg(i, i+A[i]);
		sort(seg.begin(), seg.end());
		//then process
		int reach = 0;
		for (int i = 0; i < n; ++i)
			if(seg[i].x <= reach) reach = max(reach, seg[i].y);
		if(reach >= n-1) return true;
		else return false;
	}
};

second time

class Solution {
public:
    bool canJump(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int start = 0;
        int end = 0;
        while(end < n-1)
        {
            int newEnd = start;
            for(int i = start; i <= end; ++i) newEnd = max(newEnd, i+A[i]);
            if(newEnd == end) return false;
            start = end+1;
            end = newEnd;
        }
        if(end >= n-1) return true;
        else return false;
    }
};