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MaxSubArray.java
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56 lines (52 loc) · 1.41 KB
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package array;
/**
* [53] 最大子序和
* @author: Aghao
*
* 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.
*
* 动态规划法
* 时间复杂度：O(n)
* 空间复杂度：O(1)
*
* 暴力遍历法
* 时间复杂度：O(n*n)
* 空间复杂度：O(1)
*/
public class MaxSubArray {
public static void main(String[] args) {
int[] nums = {-2,1,-3,4,-1,2,1,-5,4};
// int result = new MaxSubArray().maxSubArrayViolence(nums);
int result = new MaxSubArray().maxSubArray(nums);
System.out.println(result);
}
// 动态规划法
// 前子序列和如果小于0，就舍弃
private int maxSubArray(int[] nums) {
int pre = 0;
int maxRes = nums[0];
for(int i:nums) {
pre = Math.max(pre + i, i);
maxRes = Math.max(pre, maxRes);
}
return maxRes;
}
// 暴力法
private int maxSubArrayViolence(int[] nums) {
int maxRes = nums[0];
for(int i=0;i<nums.length;i++) {
int tempSum = 0;
for(int j=i;j<nums.length;j++) {
tempSum += nums[j];
maxRes = Math.max(maxRes, tempSum);
}
}
return maxRes;
}
}