[Vssue]0045.跳跃游戏II.md

[youngyangyang04]

https://www.programmercarl.com/0045.%E8%B7%B3%E8%B7%83%E6%B8%B8%E6%88%8FII.html

[Du1in9]

这题值得推敲！深刻诠释了贪心算法

if (nums.length == 1) return 0;
int temp = 0, cover = 0, count = 0;

for (int i = 0; i < nums.length; i++) {
    temp = Math.max(i + nums[i], temp); 
    if (i == cover) {                             
        count++;                    
        cover = temp;                  
        if (cover >= nums.length - 1) break;
    }
}
return count;

// 例: nums = [2,3,1,1,4], temp = 0, cover = 0, count = 0
i = 0: temp = 2, 满足 0 == 0, count = 1, cover = 2。没到终点, 继续遍历。
i = 1: temp = 4, 不满足 1 == 2, 继续遍历。
i = 2: temp = 4, 满足 2 == 2, count = 2, cover = 4。到终点了, 停止遍历。

[DEZREMNACUI]

这道题用动规做感觉好理解一点,但是时间复杂度就为n^2了

struct Solution;
impl Solution {
  pub fn jump(nums: Vec<i32>) -> i32 {
    let mut dp = vec![i32::MAX;nums.len()];
    dp[0] = 0;
    for i in 0..nums.len() {
      for j in 0..i {
        if nums[j] >= i as i32-j as i32{
          dp[i] = dp[i].min(dp[j]+1);
        }
      }
    }
    *dp.last().unwrap()
  }
}