maxSubArray
Description:
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
.
Thoughts:
this problem was discussed by Jon Bentley (Sep. 1984 Vol. 27 No. 9 Communications of the ACM P885)
the paragraph below was copied from his paper (with a little modifications)
algorithm that operates on arrays: it starts at the left end (element A[1]) and scans through to the right end (element A[n]), keeping track of the maximum sum subvector seen so far. The maximum is initially A[0]. Suppose we've solved the problem for A[1 .. i - 1]; how can we extend that to A[1 .. i]? The maximum
sum in the first I elements is either the maximum sum in the first i - 1 elements (which we'll call MaxSoFar), or it is that of a subvector that ends in position i (which we'll call MaxEndingHere).
MaxEndingHere is either A[i] plus the previous MaxEndingHere, or just A[i], whichever is larger.
there is my java code:
- package easy.array;
- public class MaxSubArray {
- public int maxSubArray(int[] nums){
- int maxsofar = nums[0];
- int maxtotal = nums[0];
- for(int i = 1; i< nums.length;i++){
- maxsofar = Math.max(maxsofar+nums[i], nums[i]);
- maxtotal = Math.max(maxtotal, maxsofar);
- }
- return maxtotal;
- }
- public static void main(String[] args){
- int[] nums = new int[]{-2, 1, -3, 4, -1, 2, 1, -5, 4};
- MaxSubArray max = new MaxSubArray();
- int num = max.maxSubArray(nums);
- System.out.println(num);
- }
- }
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