原题链接：

1031. 两个非重叠子数组的最大和

https://leetcode.cn/problems/maximum-sum-of-two-non-overlapping-subarrays/description/

完成情况：

解题思路：

参考代码：

package 西湖算法题解___中等题02;public class __1031两个非重叠子数组的最大和 {/**题目解释：nums[]代表数组int firstLen, int secondLen 代表两个你想要的连续子数组的长度返回最大的和，不能重叠。1 <= firstLen, secondLen <= 10002 <= firstLen + secondLen <= 1000解题思路：先找短的数组序列，再找长的数组序列。* @param nums* @param firstLen* @param secondLen* @return*/public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) {
//		int shortLen = Math.min(firstLen,secondLen);
//		int longLen = Math.max(firstLen,secondLen);
//		int n = nums.length;
//		//双重for循环，去找shortLen
//		//再双重for循环去找longLen，并且不能接触到连续的shortLen区间。int n = nums.length;int sumPrefix [] = new int[n+1];for (int i=0;i<n;i++){sumPrefix[i+1] = sumPrefix[i] + nums[i];    //计算nums的前缀和}return Math.max(findEnum(sumPrefix,firstLen,secondLen),findEnum(sumPrefix,secondLen,firstLen));}private int findEnum(int[] sumPrefix, int firstLen, int secondLen) {int maxSumA = 0,ans = 0;for (int i=firstLen + secondLen;i<sumPrefix.length;i++){maxSumA = Math.max(maxSumA,sumPrefix[i - secondLen] - sumPrefix[i-secondLen-firstLen]);ans = Math.max(ans,maxSumA+sumPrefix[i]-sumPrefix[i-secondLen]);}return ans;}
}