19.2.8 [LeetCode 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.
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.
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 };
指路→JUMP GAME II
注定失败的战争,也要拼尽全力去打赢它;
就算输,也要输得足够漂亮。