LeetCode 之 打家劫舍 I II III

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198. 打家劫舍

var rob = function(nums) {
    let n = nums.length;

    if(n === 0) {
        return 0
    }
    if (n === 1) {
        return nums[0]
    }
    
    let dp = new Array(n).fill(0);
    dp[0] = nums[0]
    dp[1] = Math.max(nums[0], nums[1])
    for (let i = 2; i < n; i++) {
        dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i])
    }
    return dp[n - 1];
};

213. 打家劫舍 II

let rob = function(nums) {
  if (nums.length === 0) return 0
  if (nums.length === 1) return nums[0]

    // 定义动态方程
  let dpHandle = function(nums, start, end) {
    if (end === start) {
        return nums[start]
    }
    const dp = new Array(nums.length).fill(0)
    dp[start] = nums[start]
    dp[start + 1] = Math.max(nums[start], nums[start + 1])
    for (let i = start + 2; i <= end; i++) {
        dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1])
    }
    return dp[end]
  }

  let ans1 = dpHandle(nums, 0, nums.length - 2)
  let ans2 = dpHandle(nums, 1, nums.length - 1)

  return Math.max(ans1, ans2)
}

337. 打家劫舍 III

const rob = root => {
  // 后序遍历函数
  const postOrder = node => {
    // 递归出口
    if (!node) return [0, 0];
    // 遍历左子树
    const left = postOrder(node.left);
    // 遍历右子树
    const right = postOrder(node.right);
    // 不偷当前节点，左右子节点都可以偷或不偷，取最大值
    const DoNot = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
    // 偷当前节点，左右子节点只能不偷
    const Do = node.val + left[0] + right[0];
    // [不偷，偷]
    return [DoNot, Do];
  };
  const res = postOrder(root);
  // 返回最大值
  return Math.max(...res);
};