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剑指 Offer 42. 连续子数组的最大和 #56

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@Geekhyt

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@Geekhyt

原题链接

状态定义

dp[i]:表示以 nums[i] 为结尾的连续子数组最大和。

状态转移方程

连续子数组则必须包含 nums[i],否则不符合题意。

dp[i - 1] > 0 时,dp[i] = dp[i - 1] + nums[i],当dp[i - 1] <= 0 时,dp[i] = nums[i]

所以状态转移方程为:

dp[i] = max(dp[i - 1] + nums[i], nums[i])

初始化

dp[0] = nums[0],以 nums[0] 为结尾的连续子数组最大和为 nums[0]。

我们只需要关注前一个状态的值,不需要额外开辟一个数组空间记录,仅用两个变量即可。

constmaxSubArray=function(nums){constlen=nums.lengthletres=nums[0]for(leti=1;i<len;i++){nums[i]=Math.max(0,nums[i-1])+nums[i]if(nums[i]>res){res=nums[i]}}returnres}
  • 时间复杂度: O(n)
  • 空间复杂度: O(1)

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