2022-06-09
Description
Given an array of non-negative integers nums
, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
You can assume that you can always reach the last index.
Example 1:
Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [2,3,0,1,4]
Output: 2
Constraints:
1 <= nums.length <= 10^4
0 <= nums[i] <= 1000
Solution
Approach #0
func jump(nums []int) (ans int) {
p, maxPos := 0, 0
for i := 0; i < len(nums)-1; i++ {
maxPos = max(maxPos, i+nums[i])
if p == i {
p = maxPos
ans++
}
}
return
}
func max(a, b int) int {
if a > b {
return a
}
return b
}45
Description
There is a robot on an m x n
grid. The robot is initially located at the top-left corner (i.e., grid[0][0]
). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]
). The robot can only move either down or right at any point in time.
Given the two integers m
and n
, return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The test cases are generated so that the answer will be less than or equal to 2 * 109
.
Example 1:
Input: m = 3, n = 7
Output: 28
Example 2:
Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
Constraints:
1 <= m, n <= 100
Solution
Approach #0
func uniquePaths(m int, n int) int {
dp := make([][]int, m)
for i := range dp {
dp[i] = make([]int, n)
dp[i][0] = 1
}
for i := 1; i < n; i++ {
dp[0][i] = 1
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
dp[i][j] = dp[i-1][j] + dp[i][j-1]
}
}
return dp[m-1][n-1]
}
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