Maximum Subarray
Given an integer array nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
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.
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 = [-2147483647] Output : -2147483647
์๋ฃ๊ตฌ์กฐ ์ ์ ํ์ ๋ณ์ 2๊ฐ ์๊ณ ๋ฆฌ์ฆ Math.max(a,b)๋ฅผ ์ด์ฉํ์ฌ ๋ ํฐ ๊ฐ์ ์ฑํํ๋ค. new๋ฅผ ํฌํจํ ์ง ๋ง์ง, new๋ถํฐ ๋ค์ ์์ํ ์ง ๋ง์ง๋ฅผ ์ ํํ๋ค.
์๊ณ ๋ฆฌ์ฆ์ Java๋ก ๊ตฌํ//๋ถํ ์ ๋ณต ์ฝ๋ ์ถ๊ฐ!
package java_basic;
public class MaxSubArray {
public static void main(String[] args) {
int[] nums = { -2, 1, -3, 4, -1, 2, 1, -5, 4 };
System.out.println(maxSubArray(nums));
}
public static int maxSubArray(int[] nums) {
int newSum = nums[0];
int max = nums[0];
for (int i = 1; i < nums.length; i++) {
newSum = Math.max(newSum + nums[i], nums[i]);
max = Math.max(max, newSum);
}
return max;
}
public static int maxSubArray_dp(int[] A) {
int n = A.length;
int[] dp = new int[n];// dp[i] means the maximum subarray ending with A[i];
dp[0] = A[0];
int max = dp[0];
for (int i = 1; i < n; i++) {
dp[i] = A[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
max = Math.max(max, dp[i]);
}
return max;
}
}
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