Maximum Subarray II
2024-08-24 02:24:27
Given an array of integers, find two non-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.
For given [1, 3, -1, 2, -1, 2], the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2], they both have the largest sum 7.
思路：遍历每一个分割线，分别求分割线左边的最大子数组和分割线右边的最大子数组。为了实现O(n) 的时间复杂度，可以创建left[] 、right[] 数组，其中left[i] 表示从0 ~ i 的数组的子数组最大值，right[j] 表示nums.length - 1 ~ j 的数组的子数组最大值。
public class Solution {
/**
* @param nums: A list of integers
* @return: An integer denotes the sum of max two non-overlapping subarrays
*/
public int maxTwoSubArrays(ArrayList<Integer> nums) {
if (nums == null || nums.size() == 0) {
return 0;
}
int[] left = new int[nums.size()];
int[] right = new int[nums.size()];
int sum = 0, max = Integer.MIN_VALUE;
for (int i = 0; i < nums.size(); i++) {
sum = sum + nums.get(i);
max = Math.max(max, sum);
sum = Math.max(sum, 0);
left[i] = max;
}
sum = 0;
max = Integer.MIN_VALUE;
for (int i = nums.size() - 1; i >= 0; i--) {
sum = sum + nums.get(i);
max = Math.max(max, sum);
sum = Math.max(sum, 0);
right[i] = max;
}
max = Integer.MIN_VALUE;
for (int i = 0; i < nums.size() - 1; i++) {
max = Math.max(left[i] + right[i + 1], max);
}
return max;
}
}
public class Solution {
/**
* @param nums: A list of integers
* @return: An integer denotes the sum of max two non-overlapping subarrays
*/
public int maxTwoSubArrays(ArrayList<Integer> nums) {
// write your code
int size = nums.size();
int[] left = new int[size];
int[] right = new int[size];
int sum = 0;
int minSum = 0;
int max = Integer.MIN_VALUE;
for(int i = 0; i < size; i++){
sum += nums.get(i);
max = Math.max(max, sum - minSum);
minSum = Math.min(sum, minSum);
left[i] = max;
}
sum = 0;
minSum = 0;
max = Integer.MIN_VALUE;
for(int i = size - 1; i >= 0; i--){
sum += nums.get(i);
max = Math.max(max, sum - minSum);
minSum = Math.min(sum, minSum);
right[i] = max;
}
max = Integer.MIN_VALUE;
for(int i = 0; i < size - 1; i++){
max = Math.max(max, left[i] + right[i + 1]);
}
return max;
}
}
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