41 - Jump Game II

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LeetCode Interview Questions 同时被 2 个专栏收录

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本文介绍了一种解决数组中从起始位置跳跃到末尾位置的算法问题，旨在找到达到最后一个元素所需的最少跳跃次数。文章详细解释了两种方法：通过设定start和end变量更新跳跃步数，并给出具体实现代码。

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.)

solution：设定start 和end ，通过遍历其间的A[i]+i值来不断更新start和end值，得到最大的A[i]+i为新end，原end+1为新的start。

class Solution {
public:
    int jump(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(A == NULL || n <= 1)
            return 0;
        
        int start = 0;;
        int end = 0;
        int step = 0;
        while(end < n)
        {
            step++;
            int max = 0;
            for(int i = start; i <= end; i++)
            {
                if(A[i] + i >= n-1)
                    return step;
                if(A[i] + i > max)
                    max = A[i] + i;
            }
            start = end + 1;
            end = max;
        }
    }
};

方法二：