53. Maximum Subarray.md

53. Maximum Subarray

Question:

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Thinking:

• Method:DP

1. 记录下每个点的值，存在二维数组中。这种方法O（On^2）,但是占空间，无法AC

class Solution {
    public int maxSubArray(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int len = nums.length;
        int[][] dp = new int[len][len];
        int result = Integer.MIN_VALUE;
        for(int i = 0; i < len; i++){
            dp[i][i] = nums[i];
            result = Math.max(result, nums[i]);
        }
        for(int i = 0; i < len - 1; i++){
            for(int j = i + 1; j < len; j++){
                dp[i][j] = dp[i][j - 1] + nums[j];
                result = Math.max(result, dp[i][j]);
            }
        }
        return result;
    }
}

• Method 2: 1.第二种方法是O（n），仍然是通过DP。
  • 缓存两个值，最大值res，以及以某个点为开头的最大值cur。
  • 什么时候更新cur,当某个点的值大于所缓存的值的时候[4, -6, 3],此时前两个的缓存值为-2，而第三个值本身为3，所以加上前两个值本身就是浪费，此时我们就以3为开头进行继续运算。

class Solution {
    public int maxSubArray(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int len = nums.length;
        int curr = nums[0];
        int res = nums[0];
        for(int i = 1; i < len; i++){
            curr = Math.max(nums[i], nums[i] + curr);
            res = Math.max(res, curr);
        }
        return res;
    }
}

二刷

二刷的时候还借鉴了一刷的结果。

class Solution {
    public int maxSubArray(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int cur = nums[0], max = nums[0];
        for(int i = 1; i < nums.length; i++){
            cur = Math.max(nums[i], cur + nums[i]);
            max = Math.max(cur, max);
        }
        return max;
    }
}

Third TIme

• Method 1:

  class Solution {
      public int maxSubArray(int[] nums) {
          int sum = nums[0], max = nums[0];
          for(int i = 1; i < nums.length; ++i){
              sum = Math.max(nums[i], sum + nums[i]);
              max = Math.max(sum, max);
          }
          return max;
      }
  }

• Method 2: dp

  • dp[i] the max value if using including current number.
  • dp[i] = dp[i - 1] > 0 ? dp[i - 1] + nums[i]: nums[i]
  • Need to use a variable to hold the global max value;

  class Solution {
      public int maxSubArray(int[] nums) {
          int[] dp = new int[nums.length];
          dp[0] = nums[0];
          int max = nums[0];
          for(int i = 1; i < nums.length; i++){
              dp[i] = dp[i - 1] > 0 ? nums[i] + dp[i - 1] : nums[i];
              max = Math.max(max, dp[i]);
          }
          return max;
      }
  }

Fourth Time

• Method 1: dp

   class Solution {
   	public int maxSubArray(int[] nums) {
   		if(nums == null || nums.length == 0) return 0;
   		int len = nums.length;
   		int res = Integer.MIN_VALUE;
   		int[] dp = new int[len + 1];
   		dp[0] = 0;
   		for(int i = 1; i <= len; i++){
   			dp[i] = Math.max(nums[i - 1], dp[i - 1] + nums[i - 1]);
   			res = Math.max(res, dp[i]);
   		}
   		return res;
   	}
   }

Amazon Session

• Method 1: dp

   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 max = nums[0];
   		for(int i = 1; i < nums.length; i++){
   			dp[i] = Math.max(nums[i] + dp[i - 1], nums[i]);
   			max = Math.max(max, dp[i]);
   		}
   		return max;
   	}
   }