时间复杂度过高，o(n2)超时了：

class Solution {public:    int jump(vector
     
      & nums) {        int length = nums.size();        if(length <= 0)            return 0;        vector
      
        result(length);        result[0] = 0;        bool flag[length];        flag[0] = true;        for(int i = 1;i < length;i++)            flag[i] = false;        for(int i = 1;i < length;i++){            int min_num = 0x7fffffff;            for(int j = i-1;j >= 0;j--){                if(flag[j] == false)                    continue;                if(flag[j] == true && nums[j] >= i-j){                    if(result[j]+1 < min_num){                        min_num = result[j] + 1;                        flag[i] = true;                    }                }            }            result[i] = min_num;        }        return result[length-1];    }};
      
     

 把上面代码简化，依旧是超时

class Solution {public:    int jump(vector
     
      & nums) {        int length = nums.size();        if(length <= 0)            return 0;        vector
      
        result(length);        result[0] = 0;        for(int i = 1;i < length;i++)            result[i] = -1;        for(int i = 1;i < length;i++){            int min_num = 0x7FFFFFFF;            for(int j = i-1;j >= 0;j--){                if(result[j] != -1 && nums[j] >= i-j){                    result[i] = result[j] + 1;                    if(result[i] < min_num)                        min_num = result[i];                }            }            if(min_num != 0x7FFFFFFF)                result[i] = min_num;            else                result[i] = -1;        }        return result[length-1];    }};
      
     

 用贪心算法做，时间复杂度只有0(n)

class Solution {public:    int jump(vector
     
      & nums) {        int n = nums.size();        if(n <= 0)            return 0;        int lastReach = 0;          int reach = 0;          int step = 0;          for(int i = 0; i < n && i <= reach; i++)          {              if(i > lastReach)              {                  step++;                  lastReach = reach;              }              reach = max(nums[i]+i, reach);          }          if(reach < n-1)              return 0;          else              return step;    }};
     

 推导过程：

两个讲解博客：

http://blog.csdn.net/cinderella_niu/article/details/42804559

http://www.cnblogs.com/boring09/p/4231771.html