lintcode题目记录4

Russian Doll Envelopes

俄罗斯玩偶嵌套问题，这个是典型的dp问题···强行遍历会提示超时，然而整了好久也没整明白怎么整，网上搜了下 把问题归结为求最长递增子序列问题··然而本人愚钝还是想不明白为啥可以这样做··虽然出来的结果是对的·····

先把数据排序， 用python内建的排序函数进行排序，但是因为当x相等时，y要按从大到小拍，所以要传一个cmp进去，python3.x不支持cmp了 所以 用了一个转换，转换成key，如果直接key设置为x默认y会按从小到大拍

这样算的结果是对的·但是那个迭代的dp不是一个有效的序列···但是长度是对的···

<span style="color: #0000ff">class</span><span style="color: #000000"> Solution:
    </span><span style="color: #008000">#</span><span style="color: #008000"> @param {int[][]} envelopes a number of envelopes with widths and heights</span>
    <span style="color: #008000">#</span><span style="color: #008000"> @return {int} the maximum number of envelopes</span>
    <span style="color: #0000ff">def</span><span style="color: #000000"> maxEnvelopes(self, envelopes):
        </span><span style="color: #008000">#</span><span style="color: #008000"> Write your code here</span>
        <span style="color: #0000ff">import</span><span style="color: #000000"> functools
        nums </span>= sorted(envelopes,key= functools.cmp_to_key(<span style="color: #0000ff">lambda</span> x,y:x[0]-y[0] <span style="color: #0000ff">if</span> x[0] != y[0] <span style="color: #0000ff">else</span> y[1] - x[1<span style="color: #000000">]))
        size </span>=<span style="color: #000000"> len(nums)
        dp </span>=<span style="color: #000000"> []
        </span><span style="color: #0000ff">for</span> x <span style="color: #0000ff">in</span><span style="color: #000000"> range(size):
            low, high </span>= 0, len(dp) - 1
            <span style="color: #0000ff">while</span> low <=<span style="color: #000000"> high:
                mid </span>= (low + high)//2
                <span style="color: #0000ff">if</span> dp[mid][1] < nums[x][1<span style="color: #000000">]:
                    low </span>= mid + 1
                <span style="color: #0000ff">else</span><span style="color: #000000">:
                    high </span>= mid - 1
            <span style="color: #0000ff">if</span> low <<span style="color: #000000"> len(dp):
                dp[low] </span>=<span style="color: #000000"> nums[x]
            </span><span style="color: #0000ff">else</span><span style="color: #000000">:
                dp.append(nums[x])
        </span><span style="color: #0000ff">return</span> len(dp)

 

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