• [leetcode]53. Maximum Subarray最大子数组和


    Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

    Example:

    Input: [-2,1,-3,4,-1,2,1,-5,4],
    Output: 6
    Explanation: [4,-1,2,1] has the largest sum = 6.

    Follow up:

    If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

    题意:

    求最大子数组和

    思路:

    开一个一维dp 

    原理:任何数加负数肯定比原值小

    1. dp[i] stands for max sum of subarray[0,i] (nums[i] must be used)

    2. coz any negative will worsen final result.  supposing I am standing at current nums[i],  if previous sum dp[i-1]  > 0, it will benifit result, then we wrap dp[i-1] + nums[i] to dp[i]

    otherwise, if dp[i-1] < 0, we only keep nums[i]

    3. the  dp function will be :

    dp[i] = dp[i-1] >0 ? dp[i-1] + num[i] : num[i]

     

    code

     1 class Solution {
     2     public int maxSubArray(int[] nums) {
     3         // dp[i]: max sum of subarray from 0 to i
     4         int[] dp = new int[nums.length];
     5         // initiliaze
     6         dp[0] = nums[0];
     7         int result = nums[0];
     8         
     9         for(int i = 1; i < nums.length; i++){
    10             // negative value will worsen result, if dp[i-1] < 0, we only keep nums[i]
    11             dp[i] = dp[i-1] > 0 ? dp[i-1] + nums[i] : nums[i];
    12             // update max result
    13             result = Math.max(result, dp[i]);         
    14         }
    15        return result; 
    16     }
    17 }

    思路

    优化空间为O(1)

    Coz result is only related to previous sum.

    We can use a variable to track previous sum.

    代码

     1 public class MaximumSubarray {
     2     public int maxSubArray(int[] nums) {
     3         // initialize
     4         int result = Integer.MIN_VALUE;
     5         // reset it to 0 if it's less than 0.
     6         int sum = 0;
     7         for (int n : nums) {
     8             // if sum < 0, we only keep n; if sum > 0, sum to be added to benefit result
     9             sum = n + Math.max(sum, 0);
    10             // update max result
    11             result = Math.max(result, sum);
    12         }
    13         return result;
    14     }
    15 }
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  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9206895.html
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