LeetCode刷题笔录Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array

[−2,1,−3,4,−1,2,1,−5,4]

,

the contiguous subarray

[4,−1,2,1]

has the largest sum
=

6

.

click to show more practice.

More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
动态规划，注意这里必须要有至少一个元素。因此条件就是对于A[i]，要么结果里有，要么没有。

public class Solution {
public int maxSubArray(int[] A) {
int[] x = new int[A.length];
x[0] = A[0];
int max = x[0];
for(int i = 1; i < A.length; i++){
x[i] = Math.max(x[i - 1] + A[i], A[i]);
max = Math.max(x[i], max);
}

return max;
}
}

Divide and conquer的做法是把数组分成左右两边，然后：

1.找左边的max subarray

2.找右边的max subarray

3.找包含中间元素的max subarray

对于1和2，用递归就行了。需要写一个函数来找3

public int maxThroughMid(int[] arr, int low, int mid, int high){
if(low == high)
return arr[low];
int leftMax = Integer.MIN_VALUE;
int sum = 0;
//find the max subarray on the left of the mid
for(int i = mid; i >= low; i--){
sum += arr[i];
leftMax = maxOfTwo(sum, leftMax);
}
//the max on the right
sum = 0;
int rightMax = Integer.MIN_VALUE;
for(int i = mid + 1; i <= high; i++){
sum += arr[i];
rightMax = maxOfTwo(sum, rightMax);
}
return leftMax + rightMax;
}

代码就是这样：

public class Solution {
public int maxSubArray(int[] A) {
return maxSubArrayRecur(A, 0, A.length - 1);
}

public int maxSubArrayRecur(int[] arr, int low, int high){
if(low == high){
return arr[low];
}
int mid = (low + high) / 2;

return maxOfThree(maxSubArrayRecur(arr, low, mid), maxSubArrayRecur(arr, low + 1, high), maxThroughMid(arr,low, mid, high));
}

public int maxOfThree(int a, int b, int c){
return maxOfTwo(maxOfTwo(a, b), c);
}

public int maxOfTwo(int a, int b){
if(a >= b)
return a;
else
return b;
}
//calculate the max subarray contanning the element arr[mid]
public int maxThroughMid(int[] arr, int low, int mid, int high){
if(low == high)
return arr[low];
int leftMax = Integer.MIN_VALUE;
int sum = 0;
//find the max subarray on the left of the mid
for(int i = mid; i >= low; i--){
sum += arr[i];
leftMax = maxOfTwo(sum, leftMax);
}
//the max on the right
sum = 0;
int rightMax = Integer.MIN_VALUE;
for(int i = mid + 1; i <= high; i++){
sum += arr[i];
rightMax = maxOfTwo(sum, rightMax);
}
return leftMax + rightMax;
}
}

但是online judge通不过，总是超时，先不管了。