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.
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.
贪心算法1:
public int maxSubArray(int[] nums) {
int sum = 0, max = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
max = Math.max(max, sum);
sum = sum < 0 ? 0 : sum;
}
return max;
}
贪心算法2:
public int maxSubArray(int[] nums) {
int sum = nums[0];
int max = nums[0];
for (int i = 1; i< nums.length; i++) {
sum = sum < 0 ? nums[i] : sum + nums[i];
max = max > sum ? max : sum;
}
return max;
}
动态规划:
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++) {
if (dp[i-1] < 0) {
dp[i] = nums[i];
} else {
dp[i] = dp[i-1] + nums[i];
}
max = Math.max(max, dp[i]);
}
return max;
}
本文介绍了使用贪心算法和动态规划求解最大连续子数组和的问题,包括两种不同的实现方式,并提供了代码示例。
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