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feat: #1031: add concise C++ implementation (azl397985856#164)

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`@@ -104,3 +104,23 @@ class Solution:`
`104`	`104`
`105`	`105`	`1. 代码中, 求解了4个动态规划数组来求解最终值, 有没有可能只用两个数组来求解该题, 可以的话, 需要保留的又是哪两个数组?`
`106`	`106`	`2. 代码中, 求解的4动态规划数组的顺序能否改变, 哪些能改, 哪些不能改?`
	`107`	`+`
	`108`	`+如果采用前缀和数组的话,可以只使用O(n)的空间来存储前缀和,O(1)的动态规划状态空间来完成。C++代码如下:`
	`109`	+```C++
	`110`	`+class Solution {`
	`111`	`+public:`
	`112`	`+ int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {`
	`113`	`+ auto tmp = vector<int>{A[0]};`
	`114`	`+ for (auto i = 1; i < A.size(); ++i) {`
	`115`	`+ tmp.push_back(A[i] + tmp[i - 1]);`
	`116`	`+ }`
	`117`	`+ auto res = tmp[L + M - 1], lMax = tmp[L - 1], mMax = tmp[M - 1];`
	`118`	`+ for (auto i = L + M; i < tmp.size(); ++i) {`
	`119`	`+ lMax = max(lMax, tmp[i - M] - tmp[i - M - L]);`
	`120`	`+ mMax = max(mMax, tmp[i - L] - tmp[i - L - M]);`
	`121`	`+ res = max(res, max(lMax + tmp[i] - tmp[i - M], mMax + tmp[i] - tmp[i - L]));`
	`122`	`+ }`
	`123`	`+ return res;`
	`124`	`+ }`
	`125`	`+};`
	`126`	+```

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