53. Maximum Subarray #
题目 #
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.
题目大意 #
给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。
解题思路 #
- 这一题可以用 DP 求解也可以不用 DP。
- 题目要求输出数组中某个区间内数字之和最大的那个值。
dp[i]表示[0,i]区间内各个子区间和的最大值，状态转移方程是dp[i] = nums[i] + dp[i-1] (dp[i-1] > 0)，dp[i] = nums[i] (dp[i-1] ≤ 0)。
代码 #
package leetcode
// 解法一 DP
func maxSubArray(nums []int) int {
if len(nums) == 0 {
return 0
}
if len(nums) == 1 {
return nums[0]
}
dp, res := make([]int, len(nums)), nums[0]
dp[0] = nums[0]
for i := 1; i < len(nums); i++ {
if dp[i-1] > 0 {
dp[i] = nums[i] + dp[i-1]
} else {
dp[i] = nums[i]
}
res = max(res, dp[i])
}
return res
}
// 解法二 模拟
func maxSubArray1(nums []int) int {
if len(nums) == 1 {
return nums[0]
}
maxSum, res, p := nums[0], 0, 0
for p < len(nums) {
res += nums[p]
if res > maxSum {
maxSum = res
}
if res < 0 {
res = 0
}
p++
}
return maxSum
}