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Ray-56 avatar Ray-56 commented on August 13, 2024
  • 设动态规划列表 dp,dp[i] 代表第 i 间房可偷到的最多现金
  • 表达式为 dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i])

复杂度为 O(N)

/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {
	if (!nums.length) return 0;
	const dp = Array(nums);
	dp[0] = nums[0];
	dp[1] = Math.max(nums[0], nums[1]);
	for (let i = 2; i < nums.length; i++) {
		dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
	}
	return dp[nums.length - 1];
};

from like-algorithms.

bazinga-web avatar bazinga-web commented on August 13, 2024
/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function (nums) {
  if (nums.length === 0) {
    return 0;
  }

  if (nums.length === 1) {
    return nums[0];
  }

  prve2 = nums[0];
  prve1 = Math.max(nums[0], nums[1]);

  for (let i = 2; i < nums.length; i++) {
    let temp = Math.max(nums[i] + prve2, prve1);
    prve2 = prve1;
    prve1 = temp;
  }
  return prve1;
};

from like-algorithms.

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