leetcode55

bool canJump(vector<int>& nums) {
    int reach = nums[];
    for (int i = ; i < nums.size() && reach >= i; i++)
    {
        if (i + nums[i] > reach)
        {
            reach = i + nums[i];  //贪心策略
        }
    }
    return reach >= (nums.size() - ) ? true : false;
}

这是贪心算法类型的题目。

补充一个python的实现：

 class Solution:
     def canJump(self, nums: 'List[int]') -> 'bool':
         n = len(nums)
         if n == 1:
             return True
         i = n - 1
         j = i - 1
         nexti = i
         while i>= 0:
             tag = False
             while j >= 0:
                 diff = i - j
                 val = nums[j]
                 if diff <= val:
                     nexti = j
                     tag = True
                 if tag:
                     i = nexti
                     j = i - 1
                     break
                 else:
                     j -= 1
             if not tag:
                 return False
             if nexti == 0:
                 return True
         return True

补充一个双指针思路的解决方案，使用python实现：

 class Solution:
     def canJump(self, nums: 'List[int]') -> bool:
         n = len(nums)
         j = #可以跳到的最远的索引
         for i in range(n):
             if i > j:#说明有无法跳跃的索引
                 return False
             j = max(j,i+nums[i])#更新最远的索引
         if j >= n - :#达到最右索引
             return True
         return False