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Leetcode – Jump Game Two

2019年04月17日 ⁄ 综合 ⁄ 共 863字 ⁄ 字号 评论关闭

和Jump Game思路差不多,都是DP,关键是要用一个数组maxNumbers[k]存储走k步的话,最远能够到达的序号,注意数组maxNumbers[]是递增的。

class Solution {
public:
	const int MAXVALUE = 1 << 30;

	int findMinStepToIndex(int maxNumbers[],int maxSteps,int index)
	{
		if (index == 0)
			return 0;

		int left = 1;
		int right = maxSteps;

		while (left < right)
		{
			int m = (left + right) / 2;

			if (maxNumbers[m] < index)
			{
				left = m + 1;
			}
			else if (maxNumbers[m] > index)
			{
				if (maxNumbers[m - 1] < index)
					return m;
				else if (maxNumbers[m - 1] == index)
					return m - 1;
				else
					right = m - 1;
			}
			else
			{
				return m;
			}
		}

		return (right + left) / 2;
	}

	int jump(int A[], int n) {
		int* maxNumbers = new int[n];//mark the max number that steps i can walk.
		int maxIndex = 0;
		int maxNumber = 0;//the max index we can walk to.
		maxNumbers[0] = 0;
		for (int i = 1; i < n; i++){
			maxNumbers[i] = 0;
		}
		int maxSteps = 0;

		for (int i = 0; i < n - 1; i++){

			if (maxNumber < i + A[i])
			{
				int cMinStep = findMinStepToIndex(maxNumbers, maxSteps, i);

				maxNumbers[cMinStep + 1] = i + A[i]; 
				maxNumber = i + A[i];

				maxSteps = cMinStep + 1;

				if (maxNumber >= n - 1)
					return maxSteps;
			}
		}
	}
};

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