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maximum subarray
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maximum-subarray/eunhwa99.java

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// 시간 복잡도: DP -> O(N)
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// 공간 복잡도: nums 배열 크기 - O(N)
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// 이전 솔루션과 동일
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// 시간 복잡도: O(n) - n은 주어진 배열의 길이
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// 공간 복잡도: O(1) - 상수 공간 사용
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class Solution {
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public int maxSubArray(int[] nums) {
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int currentSum = nums[0];
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int maxSum = currentSum;
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for (int i = 1; i < nums.length; ++i) {
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currentSum = Math.max(currentSum + nums[i], nums[i]);
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maxSum = Math.max(maxSum, currentSum);
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}
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return maxSum;
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public int maxSubArray(int[] nums) {
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int currentSum = nums[0];
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int maxSum = currentSum;
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for (int i = 1; i < nums.length; ++i) {
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currentSum = Math.max(currentSum + nums[i], nums[i]);
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maxSum = Math.max(maxSum, currentSum);
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}
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return maxSum;
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}
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}
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