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解法1 – 暴力出奇跡

窮舉出所有可能的連續(xù)子序列，并計算出它們的和，最后取它們中的最大值

空間復雜度：O(1)，時間復雜度：O （n 3）

class Solution {public int maxSubArray(int[] nums) {if (nums == null || nums.length == 0) return 0;int max = Integer.MIN_VALUE;for (int begin = 0; begin < nums.length; begin++) {for (int end = begin; end < nums.length; end++) {// sum是[begin, end]的和int sum = 0;for (int i = begin; i <= end; i++) {sum += nums[i];}max = Math.max(max, sum);}}return max;} }

所以，需要對此進行改進

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重復利用前面計算過的結果

空間復雜度：O(1)，時間復雜度：O （n 2）

class Solution {public int maxSubArray(int[] nums) {if (nums == null || nums.length == 0) return 0;int max = Integer.MIN_VALUE;for (int begin = 0; begin < nums.length; begin++) {int sum = 0;for (int end = begin; end < nums.length; end++) {// sum是[begin, end]的和sum += nums[end];max = Math.max(max, sum);}}return max;} }

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解法2 – 分治

class Solution {public int maxSubArray(int[] nums) {if (nums == null || nums.length == 0) return 0;return maxSubArray(nums, 0, nums.length);}static int maxSubArray(int[] nums, int begin, int end) {if (end - begin < 2) return nums[begin];int mid = (begin + end) >> 1;int leftMax = Integer.MIN_VALUE;int leftSum = 0;for (int i = mid - 1; i >= begin; i--) {leftSum += nums[i];leftMax = Math.max(leftMax, leftSum);}int rightMax = Integer.MIN_VALUE;int rightSum = 0;for (int i = mid; i < end; i++) {rightSum += nums[i];rightMax = Math.max(rightMax, rightSum);}return Math.max(leftMax + rightMax,Math.max(maxSubArray(nums, begin, mid),maxSubArray(nums, mid, end)));} }

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class Solution {public int maxSubArray(int[] nums) {if (nums == null || nums.length == 0) return 0;return maxSubArray(nums, 0, nums.length);}static int maxSubArray(int[] nums, int begin, int end) {if (end - begin < 2) return nums[begin];int mid = (begin + end) >> 1;int leftMax = nums[mid - 1];int leftSum = leftMax;for (int i = mid - 2; i >= begin; i--) {leftSum += nums[i];leftMax = Math.max(leftMax, leftSum);}int rightMax = nums[mid];int rightSum = rightMax;for (int i = mid + 1; i < end; i++) {rightSum += nums[i];rightMax = Math.max(rightMax, rightSum);}return Math.max(leftMax + rightMax,Math.max(maxSubArray(nums, begin, mid),maxSubArray(nums, mid, end)));} }

總結

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