[Swift]LeetCode120. 三角形最小路径和 | Triangle

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Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

---

给定一个三角形，找出自顶向下的最小路径和。每一步只能移动到下一行中相邻的结点上。

例如，给定三角形：

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。

说明：

如果你可以只使用 O(n) 的额外空间（n 为三角形的总行数）来解决这个问题，那么你的算法会很加分。

---

12ms

class Solution {
    func minimumTotal(_ triangle: [[Int]]) -> Int {
        guard triangle.count > 0 && triangle[0].count > 0 else {
            return 0
        }
        let m = triangle.count, n = triangle[0].count
        var dp = triangle[m - 1]
        
        var j = m - 2
        while j >= 0 {
            let arr = triangle[j]
            for i in 0..<arr.count {
                dp[i] = min(dp[i], dp[i + 1]) + arr[i]
            }
            
            j -= 1
        }
        
        return dp[0]
    }
}

---

16ms

class Solution {
    func minimumTotal(_ triangle: [[Int]]) -> Int {
        if triangle.count == 1 {
            return triangle[0][0]
        }
        var lastLine = triangle.last!
        var sums = Array(repeating: 0, count:triangle.count-1)
        for i in stride(from: triangle.count-2, through:0, by:-1) {
            for j in 0..<triangle[i].count {
                sums[j] = min(lastLine[j], lastLine[j+1]) + triangle[i][j]
            }
            lastLine = sums
        }
        return sums[0]
    }
}

---

44ms

class Solution {
    func minimumTotal(_ triangle: [[Int]]) -> Int {
        guard triangle.count > 0 else {
            return 0
        }
        
        var dp = triangle.last!
        
        for i in stride(from: triangle.count - 2, through: 0, by: -1) {
            for j in 0...i {
                dp[j] = min(dp[j], dp[j + 1]) + triangle[i][j]
            }
        }
        
        return dp[0]
    }
}

---

96ms

class Solution {
    
    func minimumTotal(_ triangle: [[Int]]) -> Int {
        let m = triangle.count
        if m == 0 { return 0 }
        if m == 1 { return triangle[0][0] }
        var result = triangle //设result[i][j]是到ij的最小路径和
        
        for i in 1 ..< m {
            for j in 0 ... i {
                if j == 0 {
                    result[i][j] = triangle[i][j] + result[i - 1][j]    
                } 
                else if j == i {
                    result[i][j] = triangle[i][j] + result[i - 1][j - 1]    
                }
                else {
                    result[i][j] = triangle[i][j] + min(result[i - 1][j], result[i - 1][j - 1])    
                }
            }
        }
        
        
        return result[m - 1].min() ?? 0
    }
}