[leetcode]Jump Game II

最新推荐文章于 2025-05-22 14:52:23 发布

转载最新推荐文章于 2025-05-22 14:52:23 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/lautsie/p/3254251.html

本文介绍了解决跳远问题的两种方法：一种使用了额外的数组来记录每一步的状态；另一种则更加高效，通过两个变量跟踪最大距离和当前可达最远距离，实现了在不使用额外数组的情况下解决问题。

简单题。

public class Solution {
    public int jump(int[] A) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = A.length;
        if (len == 0 || len == 1) return 0;
        
        int[] m = new int[len];
        m[0] = 0;
        int max = 0;
        int last = -1;
        for (int i = 0; i < len; i++) {
            if (i + A[i] > max) {
                last = max;
                max = i + A[i];
                
                for (int j = last+1; j <= max && j < len; j++) {
                    m[j] = m[i] + 1;
                }
            }
            
            if (max >= len - 1) return m[len-1];
        }
        return m[len-1];
    }
}

从参考的答案来看，确实可以省却m数组的：

/*
 * We use "last" to keep track of the maximum distance that has been reached
 * by using the minimum steps "ret", whereas "curr" is the maximum distance
 * that can be reached by using "ret+1" steps. Thus,
 * curr = max(i+A[i]) where 0 <= i <= last.
 */
class Solution {
public:
    int jump(int A[], int n) {
        int ret = 0;
        int last = 0;
        int curr = 0;
        for (int i = 0; i < n; ++i) {
            if (i > last) {
                last = curr;
                ++ret;
            }
            curr = max(curr, i+A[i]);
        }

        return ret;
    }
};

转载于:https://www.cnblogs.com/lautsie/p/3254251.html