55. Jump Game
Medium
You are given an integer array
nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.Return
true if you can reach the last index, or false otherwise.Example 1:
Input: nums = [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: nums = [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.
Constraints:
1 <= nums.length <= 1040 <= nums[i] <= 105
Runtime: 59 ms, faster than 95.59%
Memory Usage: 6.8 MB, less than 92.71%
func canJump(nums []int) bool {
if len(nums) <= 1 { return true }
imax := 0
for i := 0; i < len(nums) - 1; i++ {
if imax < i {
return false
}
if i + nums[i] > imax {
imax = i + nums[i]
}
if imax >= len(nums) - 1 {
return true
}
}
return false
}
DP 方法 ： 如果在第 i 格之前有格子可以跳到第 i 格，將之設為 true 。
func canJump(nums []int) bool {
dp := make([]bool, len(nums))
dp[0] = true
for i:=1; i<len(nums); i++ {
for j:=0; j<i; j++ {
if dp[j] == true && nums[j] + j >= i {
if nums[j] + j >= len(nums) - 1 {
return true
}
dp[i] = true
break
}
}
}
return dp[len(nums) - 1]
}
Last modified 6mo ago