[LeetCode] 55. Jump Game 跳跃游戏
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.
解法1：DP
解法2：贪婪算法Greedy
Java:
class Solution {
public boolean canJump(int[] A) {
int max = 0;
for(int i=0;i<A.length;i++){
if(i>max) {return false;}
max = Math.max(A[i]+i,max);
}
return true;
}
}
Python:
class Solution:
# @param A, a list of integers
# @return a boolean
def canJump(self, A):
reachable = 0
for i, length in enumerate(A):
if i > reachable:
break
reachable = max(reachable, i + length)
return reachable >= len(A) - 1
C++: Backtrack
class Solution {
public:
bool canJump(int A[], int n) {
int last=n-1,i,j;
for(i=n-2;i>=0;i--){
if(i+A[i]>=last)last=i;
}
return last<=0;
}
};
C++: DP
class Solution {
public:
bool canJump(vector<int>& nums) {
vector<int> dp(nums.size(), 0);
for (int i = 1; i < nums.size(); ++i) {
dp[i] = max(dp[i - 1], nums[i - 1]) - 1;
if (dp[i] < 0) return false;
}
return dp.back() >= 0;
}
};
C++: Greedy
class Solution {
public:
bool canJump(vector<int>& nums) {
int n = nums.size(), reach = 0;
for (int i = 0; i < n; ++i) {
if (i > reach || reach >= n - 1) break;
reach = max(reach, i + nums[i]);
}
return reach >= n - 1;
}
};
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[LeetCode] 45. Jump Game II 跳跃游戏 II
