Leetcode 55. Jump Game
2016-04-26 10:02
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问题
https://leetcode.com/problems/jump-game/解法
贪心法:now 表示要到达最后一个位置,至少要到达的位置。如果有位置i可以到达now则now = i;从后向前扫描如果now==0 则表示可以从0处到达最后一个位置。
class Solution { public: bool canJump(vector<int>& nums) { if (nums.size()==0) return false; if (nums.size() ==1) return true; int now = nums.size()-1; for (int i=nums.size()-2; i>=0; --i) { if (i+nums[i] >= now) now = i; } return now==0; } };
nums.size()-2 注意至少要有2个元素时才可以进入for 循环。
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