<p/><div class="image-package"><img src="https://upload-images.jianshu.io/upload_images/1648392-92aa45c58a77980f.jpg" contenteditable="false" img-data="{"format":"jpeg","size":104899,"height":900,"width":1600}" class="uploaded-img" style="min-height:200px;min-width:200px;" width="auto" height="auto"/>
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</p><blockquote><p>給你一個整數(shù)數(shù)組 nums 和一個整數(shù) k 忘巧，找出三個長度為 k 玄捕、互不重疊、且 3 * k 項的和最大的子數(shù)組，并返回這三個子數(shù)組禾嫉。

以下標的數(shù)組形式返回結果灾杰，數(shù)組中的每一項分別指示每個子數(shù)組的起始位置（下標從 0 開始）。如果有多個結果熙参，返回字典序最小的一個艳吠。

示例 1：

輸入：nums = [1,2,1,2,6,7,5,1], k = 2
輸出：[0,3,5]
解釋：子數(shù)組 [1, 2], [2, 6], [7, 5] 對應的起始下標為 [0, 3, 5]。
也可以取 [2, 1], 但是結果 [1, 3, 5] 在字典序上更大孽椰。
示例 2：

輸入：nums = [1,2,1,2,1,2,1,2,1], k = 2
輸出：[0,2,4]

提示：

1 <= nums.length <= 2 * 104
1 <= nums[i] < 216
1 <= k <= floor(nums.length / 3)

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/maximum-sum-of-3-non-overlapping-subarrays
著作權歸領扣網(wǎng)絡所有昭娩。商業(yè)轉載請聯(lián)系官方授權，非商業(yè)轉載請注明出處黍匾。</p></blockquote><h1 id="7obdf">題解</h1><div class="image-package"><img src="https://upload-images.jianshu.io/upload_images/1648392-452e9b7ffd18ecbf.jpg" contenteditable="false" img-data="{"format":"jpeg","size":23601,"height":381,"width":750}" class="uploaded-img" style="min-height:200px;min-width:200px;" width="auto" height="auto"/>
</div><h2 id="dhwj8">Swift</h2><blockquote><p><b/><span style="font-family: -apple-system, BlinkMacSystemFont, "Segoe UI", "PingFang SC", "Hiragino Sans GB", "Microsoft YaHei", "Helvetica Neue", Helvetica, Arial, sans-serif, "Apple Color Emoji", "Segoe UI Emoji", "Segoe UI Symbol"; font-size: 14px;"/>
</p>class Solution {
func maxSumOfThreeSubarrays(_ nums: [Int], _ k: Int) -> [Int] {
var ans = [Int](repeating: 0, count: 3)

var sum1 = 0, maxSum1 = 0, maxSum1Idx = 0
var sum2 = 0, maxSum2 = 0, maxSum2Idx1 = 0, maxSum2Idx2 = 0
var sum3 = 0, maxTotal = 0

for i in (k * 2) ..< nums.count {
sum1 += nums[i - k * 2]
sum2 += nums[i - k]
sum3 += nums[i]

if i >= k * 3 - 1 {
if sum1 > maxSum1 {
maxSum1 = sum1
maxSum1Idx = i - k * 3 + 1
}
if maxSum1 + sum2 > maxSum2 {
maxSum2 = maxSum1 + sum2
maxSum2Idx1 = maxSum1Idx
maxSum2Idx2 = i - k * 2 + 1
}
if maxSum2 + sum3 > maxTotal {
maxTotal = maxSum2 + sum3
ans[0] = maxSum2Idx1
ans[1] = maxSum2Idx2
ans[2] = i - k + 1
}

sum1 -= nums[i - k * 3 + 1]
sum2 -= nums[i - k * 2 + 1]
sum3 -= nums[i - k + 1]
}
}

return ans
}
}

print(Solution().maxSumOfThreeSubarrays([1, 2, 1, 2, 6, 7, 5, 1], 2))
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