Given an integer array nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example 1:
Input: nums = [-2,1,-3,4,-1,2,1,-5,4] Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
Example 2:
Input: nums = [1] Output: 1
Example 3:
Input: nums = [0] Output: 0
Example 4:
Input: nums = [-1] Output: -1
Example 5:
Input: nums = [-100000] Output: -100000
Constraints:
1 <= nums.length <= 3 * 104
-105 <= nums[i] <= 105
Follow up: If you have figured out the O(n)
solution, try coding another solution using the divide and conquer approach, which is more subtle.
最大子数组。
题意是给一个数组,请找出其sum值最大的子数组,只要求找到这个sum值即可。
思路是动态规划。设 dp[i] 是以 nums[i] 结尾的子数组的和。扫描数组,当遇到某个数i的时候,需要判断 dp[i - 1] 是否小于0。如果小于0,nums[i] + dp[i - 1] 的结果只会拖累 dp[i];如果大于0,可以将 dp[i] = dp[i - 1] + nums[i] 。最后return最大的 dp[i] 即可。
时间O(n)
空间O(n)
JavaScript实现
1 /** 2 * @param {number[]} nums 3 * @return {number} 4 */ 5 var maxSubArray = function(nums) { 6 let dp = new Array(nums.length); 7 dp[0] = nums[0]; 8 let res = nums[0]; 9 for (let i = 1; i < nums.length; i++) { 10 dp[i] = dp[i - 1] < 0 ? nums[i] : dp[i - 1] + nums[i]; 11 res = Math.max(res, dp[i]); 12 } 13 return res; 14 };
Java实现
1 class Solution { 2 public int maxSubArray(int[] nums) { 3 int[] dp = new int[nums.length]; 4 dp[0] = nums[0]; 5 int res = nums[0]; 6 for (int i = 1; i < nums.length; i++) { 7 dp[i] = nums[i] + (dp[i - 1] < 0 ? 0 : dp[i - 1]); 8 res = Math.max(res, dp[i]); 9 } 10 return res; 11 } 12 }
这种DP的思路也有节省空间的做法,其实我们并不一定需要知道每个dp[i]值的大小,我们只需要在遍历过程中记录一下最大的dp值即可。思路也是很类似,如果之前的值prev小于0,一定会拖累cur的,所以cur = nums[i];反之如果prev大于0,cur就变为nums[i] + prev。每次用res记录一下当前的子数组的最大值之后,就可以把cur赋给prev以达到节省空间的目的了。
时间O(n)
空间O(1)
1 class Solution { 2 public int maxSubArray(int[] nums) { 3 int prev = nums[0]; 4 int cur = nums[0]; 5 int res = nums[0]; 6 for (int i = 1; i < nums.length; i++) { 7 if (prev < 0) { 8 cur = nums[i]; 9 } else { 10 cur = prev + nums[i]; 11 } 12 res = Math.max(res, cur); 13 prev = cur; 14 } 15 return res; 16 } 17 }
还有另外一种思路,类似DP的做法,是去求局部最大和整体最大。设两个变量sum和res,sum记录当遍历到nums[i]的时候,sum + nums[i]是否对值的扩大有帮助;res记录最大的sum值。
时间O(n)
空间O(1)
Java实现
1 class Solution { 2 public int maxSubArray(int[] nums) { 3 int res = nums[0]; 4 int sum = nums[0]; 5 for (int i = 1; i < nums.length; i++) { 6 sum = Math.max(nums[i], sum + nums[i]); 7 res = Math.max(res, sum); 8 } 9 return res; 10 } 11 }
JavaScript实现
1 /** 2 * @param {number[]} nums 3 * @return {number} 4 */ 5 var maxSubArray = function(nums) { 6 let res = nums[0]; 7 let sum = nums[0]; 8 for (let i = 1; i < nums.length; i++) { 9 sum = Math.max(nums[i], sum + nums[i]); 10 res = Math.max(res, sum); 11 } 12 return res; 13 };
相关题目
918. Maximum Sum Circular Subarray