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Add Go Solution
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@ -116,8 +116,33 @@ Python:
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Go:
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```
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func lengthOfLIS(nums []int ) int {
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dp := []int{}
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for _, num := range nums {
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if len(dp) ==0 || dp[len(dp) - 1] < num {
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dp = append(dp, num)
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} else {
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l, r := 0, len(dp) - 1
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pos := r
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for l <= r {
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mid := (l + r) >> 1
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if dp[mid] >= num {
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pos = mid;
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r = mid - 1
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} else {
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l = mid + 1
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}
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}
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dp[pos] = num
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}//二分查找
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}
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return len(dp)
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}
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```
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复杂度分析
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- 时间复杂度:O(nlogn)。数组 nums 的长度为 n,我们依次用数组中的元素去更新 d 数组,而更新 d 数组时需要进行 O(logn) 的二分搜索,所以总时间复杂度为 O(nlogn)。
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- 空间复杂度:O(n),需要额外使用长度为 n 的 d 数组。
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-----------------------
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