【分治】 【动态规划】
给定一个整数数组 nums ,找到一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。
Example:
输入: [-2,1,-3,4,-1,2,1,-5,4],
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大,为 6。
Solutions
Solution ( 15ms)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
|
class Solution {
public int maxSubArray(int[] nums) {
int res = nums[0];
int sum = res;
for (int i = 1, length = nums.length; i < length; i++) {
if (sum > 0) {
sum += nums[i];
} else {
sum = nums[i];
}
res = Math.max(res, sum);
}
return res;
}
}
|
复杂度分析 6N
Solution ( 7ms)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
|
class Solution {
public int maxSubArray(int[] nums) {
int maxSum = nums[0];
int curSum = 0;
for (int n: nums) {
curSum += n;
if (curSum > maxSum) { maxSum = curSum; }
if (curSum < 0) { curSum = 0; }
}
return maxSum;
}
}
|
复杂度分析 4N
Solution 【分治】【动态规划】(9ms)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
|
class Solution {
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int[] dp = new int[nums.length];
dp[0] = nums[0];
int res = dp[0];
for (int i = 1; i < nums.length; i++) {
dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);
res = Math.max(dp[i], res);
}
return res;
}
}
|