Jump Game II

 

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.)

 

先弄了个一维DP，小数据倒是过了，但是大数据超时。

然后想复杂了，心想用线段树来保存每个数的跳跃区间的最小跳，来简化查找，这样变成 n logn 的复杂度，后来网上搜了一下，发现原来简单的贪心就可以了，自己还是太搓啊！

 

class Solution {
 private:
     int f[100000];
 public:
     int jump(int A[], int n) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         int maxPos = 0;
         f[0] = 0;
         
         for(int i = 0; i <= maxPos; i++)
         {
             int pos = i + A[i];
             if (pos >= n)
                 pos = n - 1;
                 
             if (pos > maxPos)
             {
                 for(int j = maxPos + 1; j <= pos; j++)
                     f[j] = f[i] + 1;
                 maxPos = pos; 
             }
             
             if (maxPos == n - 1)
                 return f[n-1];
         }
     }
 };