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| 1 | +# 120. 三角形最小路径和 |
| 2 | + |
| 3 | +[url](https://leetcode-cn.com/problems/triangle/) |
| 4 | + |
| 5 | + |
| 6 | +## 题目 |
| 7 | + |
| 8 | +给定一个三角形 `triangle` ，找出自顶向下的最小路径和。 |
| 9 | + |
| 10 | +每一步只能移动到下一行中相邻的结点上。相邻的结点 在这里指的是 下标 与 上一层结点下标 相同或者等于 上一层结点下标 + 1 的两个结点。也就是说，如果正位于当前行的下标 i ，那么下一步可以移动到下一行的下标 i 或 i + 1 。 |
| 11 | + |
| 12 | + |
| 13 | +``` |
| 14 | +输入：triangle = [[2],[3,4],[6,5,7],[4,1,8,3]] |
| 15 | +输出：11 |
| 16 | +解释：如下面简图所示： |
| 17 | + 2 |
| 18 | + 3 4 |
| 19 | + 6 5 7 |
| 20 | +4 1 8 3 |
| 21 | +自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。 |
| 22 | +``` |
| 23 | + |
| 24 | +## 方法 |
| 25 | + |
| 26 | +## code |
| 27 | + |
| 28 | +### js |
| 29 | + |
| 30 | +```js |
| 31 | +let minimumTotal = triangle => { |
| 32 | + if (triangle.length === 0) |
| 33 | + return 0; |
| 34 | + let row = triangle.length; |
| 35 | + let dp = Array(row).fill(0).map(_ => Array(triangle[row-1].length).fill(0)); |
| 36 | + // 初始化 |
| 37 | + for (let i = 0; i < row; i++) { |
| 38 | + for (let j = 0; j < triangle[i].length; j++) { |
| 39 | + dp[i][j] = triangle[i][j]; |
| 40 | + } |
| 41 | + } |
| 42 | + // 从下往上，初始化最后一行 |
| 43 | + for (let i = 0; i < triangle[row-1].length; i++) { |
| 44 | + dp[row-1][i] = triangle[row-1][i]; |
| 45 | + } |
| 46 | + // dp |
| 47 | + for (let i = row-2; i >= 0; i--) { |
| 48 | + for (let j = 0; j < triangle[i].length; j++) { |
| 49 | + dp[i][j] = Math.max(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j]; |
| 50 | + } |
| 51 | + } |
| 52 | + return dp[0][0]; |
| 53 | +} |
| 54 | +``` |
| 55 | + |
| 56 | +### go |
| 57 | + |
| 58 | +```go |
| 59 | +func minimumTotal(triangle [][]int) int { |
| 60 | + Min := func(a, b int) int { |
| 61 | + if a < b { |
| 62 | + return a |
| 63 | + } else { |
| 64 | + return b |
| 65 | + } |
| 66 | + } |
| 67 | + if len(triangle) == 0 { |
| 68 | + return 0 |
| 69 | + } |
| 70 | + row := len(triangle) |
| 71 | + dp := make([][]int, row) |
| 72 | + for i := range dp { |
| 73 | + dp[i] = make([]int, len(triangle[row-1])) |
| 74 | + } |
| 75 | + // 初始化 |
| 76 | + for i := 0; i < row; i++ { |
| 77 | + for j := 0; j < len(triangle[i]); j++ { |
| 78 | + dp[i][j] = triangle[i][j] |
| 79 | + } |
| 80 | + } |
| 81 | + // 从下往上，初始化最后一行 |
| 82 | + for i := 0; i < len(triangle[row-1]); i++ { |
| 83 | + dp[row-1][i] = triangle[row-1][i] |
| 84 | + } |
| 85 | + // dp |
| 86 | + for i := row-2; i >= 0; i-- { |
| 87 | + for j := 0; j < len(triangle[i]); j++ { |
| 88 | + dp[i][j] = Min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] |
| 89 | + } |
| 90 | + } |
| 91 | + return dp[0][0] |
| 92 | +} |
| 93 | +``` |
| 94 | + |
| 95 | +### java |
| 96 | + |
| 97 | +```java |
| 98 | +class Solution { |
| 99 | + public int minimumTotal(List<List<Integer>> triangle) { |
| 100 | + if(triangle.size() == 0) return 0; |
| 101 | + int row = triangle.size(); |
| 102 | + int[][] dp = new int[row][triangle.get(row - 1).size()]; |
| 103 | + // 初始化 |
| 104 | + for(int i = 0; i < row; i++) { |
| 105 | + for (int j =0; j < triangle.get(i).size(); j++) { |
| 106 | + dp[i][j] = triangle.get(i).get(j); |
| 107 | + } |
| 108 | + } |
| 109 | + // 从下往上， 初始化最后一行 |
| 110 | + for (int i = 0; i < triangle.get(row - 1).size(); i++) { |
| 111 | + dp[row - 1][i] = triangle.get(row - 1).get(i); |
| 112 | + } |
| 113 | + // 动态规划 |
| 114 | + for (int i = row - 2; i >= 0; i--) { |
| 115 | + for (int j = 0; j < triangle.get(i).size(); j++) { |
| 116 | + dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + triangle.get(i).get(j); |
| 117 | + } |
| 118 | + } |
| 119 | + return dp[0][0]; |
| 120 | + } |
| 121 | +} |
| 122 | +``` |
| 123 | + |
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