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19.2.8 [LeetCode 55] Jump Game

时间:2019-02-08 18:05:51      阅读:195      评论:0      收藏:0      [点我收藏+]

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.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.
技术分享图片
 1 class Solution {
 2 public:
 3     bool canJump(vector<int>& nums) {
 4         if (nums.size() == 1)return true;
 5         int reach = nums[0], prereach = 0, n = nums.size();
 6         while (1) {
 7             int nextreach = reach;
 8             bool flag = false;
 9             for (int i = prereach + 1; i <= reach && i < n; i++) {
10                 if (i + nums[i] > nextreach) {
11                     nextreach = i + nums[i];
12                     flag = true;
13                 }
14             }
15             if (nextreach >= n - 1)
16                 return true;
17             if (!flag)return false;
18             prereach = reach;
19             reach = nextreach;
20         }
21         return false;
22     }
23 };
View Code

指路→JUMP GAME II

19.2.8 [LeetCode 55] Jump Game

原文:https://www.cnblogs.com/yalphait/p/10356443.html

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