LeetCode 笔记系列13 Jump Game II [去掉不必要的计算]

题目： 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.)

当本娃拿到这个题目的时候，第一反应必然是dp。

解法一：

 1  public static int jump(int[] A) {
 2             // Start typing your Java solution below
 3             // DO NOT write main() function
 4          if(A.length < 2) return 0;
 5          int[] dist = new int[A.length];
 6          int[] to = new int[A.length];
 7          for(int i = 0; i < A.length; i++){
 8              dist[i] = INFINITE;
 9          }
10          dist[A.length - 1] = 0;
11          for(int i = A.length - 2; i >= 0; i--){
12              int minDist = INFINITE;
13              for(int j = 1; j <= A[i] && i + j < A.length; j++){
14                  int nextIdx = i + j;
15                  if(nextIdx < A.length){
16                      int candidate = dist[nextIdx] + 1;
17                      if(candidate < minDist){
18                          minDist = candidate;
19                          to[i] = nextIdx;
20                      }
21                  }
22              }
23              dist[i] = minDist;
24          }
25          return dist[0];
26      }

View Code

然后，提交，大集合再次不过。。。WTF。。。

左思右想不得其解。

好在有leetcode讨论版，看到大牛的一个几行的代码。

NB闪闪的，就不折叠鸟。

解法二：

/*
 * 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;
    }
};