最小子列和的动态规划解法

最小子列和的动态规划

53. Maximum Subarray

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.

分治法：

public int maxSubArray(int[] nums) {
        return devideconquer(nums,0,nums.length-1);
    }
    public int devideconquer(int[] nums, int l, int r){
        if(l==r) return nums[l];
        int mid = (l+r)/2;
        int maxLeft = devideconquer(nums,l,mid);
        int maxRight = devideconquer(nums,mid+1,r);
        int maxSide = Math.max(maxLeft,maxRight);
        
        int acrossLeft = Integer.MIN_VALUE;
        int sumLeft = 0;
        for(int i=mid;i>=l;i--){
            sumLeft += nums[i];
            acrossLeft = Math.max(acrossLeft,sumLeft);
        }
        
        int acrossRight = Integer.MIN_VALUE;
        int sumRight = 0;
        for(int i=mid+1;i<=r;i++){
            sumRight += nums[i];
            acrossRight = Math.max(acrossRight,sumRight);
        }
        
        return Math.max(maxSide,acrossLeft+acrossRight);
    }

动态规划法：

若加和为正数则会继续向后加，若加和为负数，则停止又从下一个数开始。

public int maxSubArray(int[] nums) {
        int[] dp=new int[nums.length];
        dp[0]=nums[0];
        int Max=nums[0];
        for(int i=1;i<dp.length;i++){
            dp[i]=Math.max(dp[i-1]+nums[i],nums[i]);
            Max=Math.max(dp[i],Max);
        }
        return Max;
    }