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Copy pathminimum-sideway-jumps.rs
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minimum-sideway-jumps.rs
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#![allow(dead_code, unused, unused_variables, non_snake_case)]
fn main() {
assert_eq!(Solution::min_side_jumps(vec![0, 2, 1, 0, 3, 0]), 2);
assert_eq!(Solution::min_side_jumps(vec![0, 1, 1, 3, 3, 0]), 0);
assert_eq!(Solution::min_side_jumps(vec![0, 1, 2, 3, 0]), 2);
}
struct Solution;
impl Solution {
/// dp 这个看不懂
pub fn min_side_jumps1(obstacles: Vec<i32>) -> i32 {
let mut dp = [1, 0, 1i64];
// 起点都没障碍物,所以起点为0
for i in 1..obstacles.len() {
let a = obstacles[i];
// 说明当前位置三个跑道都没有障碍物,因此
let mut new_dp = [0, 0, 0];
if a == 0 {
new_dp[0] = dp[0].min(1 + dp[1].min(dp[2]));
new_dp[1] = dp[1].min(1 + dp[0].min(dp[2]));
new_dp[2] = dp[2].min(1 + dp[0].min(dp[1]));
} else if a == 1 {
new_dp[0] = i32::MAX as i64;
new_dp[1] = dp[1].min(1 + dp[2]);
new_dp[2] = dp[2].min(1 + dp[1]);
} else if a == 2 {
new_dp[0] = dp[0].min(1 + dp[2]);
new_dp[1] = i32::MAX as i64;
new_dp[2] = dp[2].min(1 + dp[0]);
} else {
new_dp[0] = dp[0].min(1 + dp[1]);
new_dp[1] = dp[1].min(1 + dp[0]);
new_dp[2] = i32::MAX as i64;
}
dp = new_dp;
}
*dp[..].into_iter().min().unwrap() as i32
}
/// dp[i][j], 0 < i < 3, j < obstacles.len() 表示第i条跑道在第j点时的最少跳跃次数
/// 所以dp[i][j] 可能有:1.如果有障碍物的话,就不符合要求
/// 2.没有障碍物,就是dp[i-1][j]或者是dp[i][n]+1的最小值(因为跳跃到[i,j]位置不值可以从[i-1,j]跳跃而来,而且还可以从j处的其他跑道而来。所以统一下跳跃的次数就行了。)
pub fn min_side_jumps(obstacles: Vec<i32>) -> i32 {
// 起点都没障碍物,所以起点为0
let mut dp = [1, 0, 1i64];
for i in 1..obstacles.len() {
let a = obstacles[i]; // 障碍物
let mut new_dp = [0, 0, 0];
let mut min = i32::MAX as i64;
for i in 0..3 {
if a - 1 == i {
new_dp[i as usize] = i32::MAX as i64;
} else {
new_dp[i as usize] = dp[i as usize];
}
min = min.min(new_dp[i as usize]);
}
// 因为任何点都可以从当前位置的其他点+1跳来,需要判断下是不是其他点+1跳过来的更短少
for i in 0..3 {
if a - 1 != i {
new_dp[i as usize] = new_dp[i as usize].min(min + 1);
}
}
dp = new_dp;
}
*dp[..].into_iter().min().unwrap() as i32
}
}