描述
给定一个整数数组,找出两个不重叠的子数组A和B,使两个子数组和的差的绝对值|SUM(A) - SUM(B)|最大。
返回这个最大的差值。
子数组最少包含一个数
样例
给出数组[1, 2, -3, 1],返回 6
挑战
时间复杂度为O(n),空间复杂度为O(n)
思路:与上一题最大子数组II类似,只是要分情况找到sum(A)最大,sum(B)最小和sum(A)最小,sum(B)最大的情况
public int maxDiffSubArrays(int[] nums) {
// write your code here
int[] leftMax = new int[nums.length];
int[] leftMin = new int[nums.length];
int[] rightMin = new int[nums.length];
int[] rightMax = new int[nums.length];
int curSum = 0;
int maxSum = nums[0];
int curDiff = 0;
int minSum = nums[0];
for (int i = 0; i < nums.length; i++) {
curSum += nums[i];
maxSum = Math.max(maxSum, curSum);
curSum = Math.max(curSum, 0);
leftMax[i] = maxSum;
curDiff += nums[i];
minSum = Math.min(minSum, curDiff);
curDiff = Math.min(curDiff, 0);
leftMin[i] = minSum;
}
curSum = 0;
curDiff = 0;
maxSum = nums[nums.length - 1];
minSum = nums[nums.length - 1];
for (int i = nums.length - 1; i >= 0; i--) {
curDiff += nums[i];
minSum = Math.min(minSum, curDiff);
curDiff = Math.min(curDiff, 0);
rightMin[i] = minSum;
curSum += nums[i];
maxSum = Math.max(maxSum, curSum);
curSum = Math.max(curSum, 0);
rightMax[i] = maxSum;
}
int result = 0;
for (int i = 0; i < nums.length - 1; i++) {
result = Math.max(result, Math.abs(leftMax[i] - rightMin[i + 1]));
result = Math.max(result, Math.abs(leftMin[i] - rightMax[i + 1]));
}
return result;
}