【Leetcode】Jump Game
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55. Jump Game
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.
Determine if you are able to reach the last index.
For example:
A = [2,3,1,1,4], return true.
A = [3,2,1,0,4], return false.
题目:给定一组非负的整数,初始时在数组中的第一个元素的位置。元素中的每一个元素表示当前位置可以跳的最大步数。需要决定是否能够达到最后一个位置。
例如,给定A=[2,3,1,1,4],返回真。给定A=[3,2,1,0,4],返回真。
解题思路
这道题目其实可以直接用动态规划:
设状态为f[i],表示从第0 层出发,走到A[i] 时剩余的最大步数,则状态转移方程为:
f[i] = max(f[i - 1], A[i - 1]) - 1, i > 0
class Solution {public: bool canJump(vector<int>& nums) { int n = nums.size(); vector<int> f(n, 0); f[0] = 0; for (int i = 1; i < n; i++) { f[i] = max(f[i - 1], nums[i - 1]) - 1; if (f[i] < 0) return false;; } return f[n - 1] >= 0; }};
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