一天一大 leet(三角形最小路径和)难度:中等-Day20200714

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

[
  [2,0,0,0],
  [3,4,0,0],
  [6,5,7,0],
  [4,1,8,3]
]
=>
[
  [2,   0,        0,    0],
  [5,   6,        0,    0],
  [11,  (10,11),  13,   (0,6)],
  [...]
]

从左上角到右下角

/**
 * @param {number[][]} triangle
 * @return {number}
 */
var minimumTotal = function (triangle) {
  let n = triangle.length,
    dp = Array(n)
  // 声明nXn数组存放结果
  for (let i = 0; i < n; i++) {
    dp[i] = []
  }
  // 放置初始值(第一步累加)
  dp[0][0] = triangle[0][0]

  for (let i = 1; i < n; ++i) {
    // 左上角(与其累加的只有行i不同下标一致0)
    dp[i][0] = dp[i - 1][0] + triangle[i][0]
    // 行内累加
    for (let j = 1; j < i; ++j) {
      // 下一次下标(i)(j)相等 下一次下标减一(i)(j-1) 中去最小
      dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]) + triangle[i][j]
    }
    // 右下角(与其累加只有上一行i-1下标减1，j-1)
    dp[i][i] = dp[i - 1][i - 1] + triangle[i][i]
  }
  let _result = dp[n - 1][0]
  // 取最后一行最小值
  for (let i = 1; i < n; ++i) {
    _result = Math.min(_result, dp[n - 1][i])
  }
  return _result
}

/**
 * @param {number[][]} triangle
 * @return {number}
 */
var minimumTotal = function (triangle) {
  let n = triangle.length;
  let dp = Array(n);
  for (let i = 0; i < n; i++) {
    dp[i] = [];
  }
  for (let i = n - 1; i >= 0; i--) {
    for (let j = triangle[i].length - 1; j >= 0; j--) {
      if (i == n - 1) {
        dp[i][j] = triangle[i][j];
      } else {
        dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle[i][j];
      }
    }
  }
  return dp[0][0];
}

/**
 * @param {number[][]} triangle
 * @return {number}
 */
var minimumTotal = function (triangle) {
    let dp = triangle
    for (let i = dp.length - 2; i >= 0; i--) {
        for (let j = 0; j < dp[i].length; j++) {
            dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + dp[i][j];
        }
    }
    return dp[0][0]
}

/**
 * @param {number[][]} triangle
 * @return {number}
 */
var minimumTotal = function (triangle) {
  let n = triangle.length;
  let dp = Array(triangle[n - 1].length)
  for (let i = 0; i < dp.length; i++) {
    dp[i] = triangle[n - 1][i]
  }
  for (let i = dp.length - 2; i >= 0; i--) {
    for (let j = 0; j < triangle[i].length; j++) {
      dp[j] = Math.min(dp[j], dp[j + 1]) + triangle[i][j]
    }
  }
  return dp[0]
}

/**
 * @param {number[][]} triangle
 * @return {number}
 */
var minimumTotal = function (triangle) {

  function dfs(triangle,i,j){
    if(i === triangle.length){
      return 0
    }
    return Math.min(dfs(triangle,i + 1,j),dfs(triangle,i + 1,j + 1)) + triangle[i][j]
  }

  return  dfs(triangle, 0, 0);
}