leetcode1031

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + … + A[i+L-1]) + (A[j] + A[j+1] + … + A[j+M-1]) and either:

0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
0 <= j < j + M - 1 < i < i + L - 1 < A.length.

主要是预处理
s[i] 前i个和
l[i] 前i个中,长度为L的连续子数组和
m[i] 前i个中,长度为M的连续子数组和

f[i] = max(f[i-1],s[i]-s[i-L]+m[i-L],s[i]-s[i-M]+l[i-M])

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        int len = A.length;
        int[] l = new int[len];
        int[] m = new int[len];
        int[] s = new int[len];
        int[] f = new int[len];
        s[0] = A[0];
        for(int i=1;i<len;i++){
            s[i] = s[i-1]+A[i];
        }
        l[L-1] = s[L-1];
        for(int i=L;i<len;i++){
            l[i] = max(s[i]-s[i-L],l[i-1]);
        }
        m[M-1] = s[M-1];
        for(int i=M;i<len;i++){
            m[i] = max(s[i]-s[i-M],m[i-1]);
        }
        int t = s[L+M-1];
        for(int i=M+L;i<len;i++){
            t = max(t,max(s[i]-s[i-L]+m[i-L],s[i]-s[i-M]+l[i-M]));
        }
        return t;     
    }
    
    private int max(int a,int b){
        if (a>b){
            return a;
        }
        return b;
    }
}

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