題目：

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.

思路：

保存兩個值max和leftSum，max就是到目前為止所能得到的最大值，leftSum為包含當前元素的最大子串之和。然后從左往右掃，下面的代碼的if else語句把各種情況都擺明了。

package array;public class MaximumSubarray {public int maxSubArray(int[] nums) {int n;if (nums == null || (n = nums.length) == 0) return 0;int max = nums[0];int leftSum = nums[0];for (int i = 1; i < n; ++i) {if (nums[i] >= 0) {if (leftSum >= 0) {leftSum += nums[i]; } else {leftSum = nums[i];}if (max < leftSum)max = leftSum;} else {if (leftSum + nums[i] >= 0) {leftSum += nums[i];} else {leftSum = nums[i];}if (max < leftSum)max = leftSum;}}return max;}public static void main(String[] args) {// TODO Auto-generated method stubint[] nums = { /*-2,1,-3,4,-1,2,1,-5,4*/ -2, -1 };MaximumSubarray m = new MaximumSubarray();System.out.println(m.maxSubArray(nums));}}

這個代碼可以合并，如下：

package array;public class MaximumSubarray {public int maxSubArray(int[] nums) {int n;if (nums == null || (n = nums.length) == 0) return 0;int max = nums[0];int leftSum = nums[0];for (int i = 1; i < n; ++i) {if (leftSum >= 0 && leftSum + nums[i] >= 0) {leftSum += nums[i];} else {leftSum = nums[i];}if (max < leftSum)max = leftSum;}return max;}public static void main(String[] args) {// TODO Auto-generated method stubint[] nums = { -2, -1 };MaximumSubarray m = new MaximumSubarray();System.out.println(m.maxSubArray(nums));}}

?

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