leetcode55

bool canJump(vector<int>& nums) {
    int reach = nums[0];
    for (int i = 1; i < nums.size() && reach >= i; i++)
    {
        if (i + nums[i] > reach)
        {
            reach = i + nums[i];  //贪心策略
        }
    }
    return reach >= (nums.size() - 1) ? 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 = 0#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 - 1:#达到最右索引
            return True
        return False