Leetcode-candy
2014-11-21 11:28
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题目内容:
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
keypoint: 每个孩子最少有一颗糖,多rating的孩子糖比邻居多,最少化糖数值。
思路:求最少,最优,主流思想还是贪心或者动态规划,这里增加孩子数目其rating与前面旅清还是有点困难,因此,可考虑贪心。
贪心的话,首先要保证每个孩子有1颗糖,其次,把不“和谐”的关系修理就OK。
考虑从左到右,多次修理,算法复杂度O(N*N)。
while(1){ bool te = false;for(unsigned i=1;i<m;++i){ if((ratings[i] > ratings[i-1]) && (x[i] <= x[i-1])){ x[i] = x[i-1] + 1; te = true; } else if((ratings[i] < ratings[i-1]) && (x[i] >= x[i-1])){ x[i-1] = x[i] + 1; te = true; }} if(!te) break;}一次从左至右的扫描修理,只处理了右-左正常,可能的情况是,修理之后,下一个位置会把这种合理性修改掉。如下面例子。
1 3 4 5 4 3 2 1
1 1 1 1 1 1 1 1
1 2 3 4 2 2 2 1 first process left to right
1 2 3 4 3 3 2 1 second process left to right
存在一个2-1后2-2的不合理情况,即只剩下先左后右不合理分支。处理掉这种分支,可以考虑反过来修理。(因为,反过来修理会提前把后面出现的不合理逐一转换为合理)
左右修理:(左-右)合理-合理....合理-(处理本位置使之合理,但前者就不合理了,即最后的肯定是合理的,多次循环还是可能会有合理变不合理)
右左修理:(右-左)既然最后合理,那么反过来修理,这个不合理列最右边肯定是2 1之流(连续减小才会碰到左-右修理搞不定),即最右为正确数值(value是正常),往左刷一次即可完成修理。
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
keypoint: 每个孩子最少有一颗糖,多rating的孩子糖比邻居多,最少化糖数值。
思路:求最少,最优,主流思想还是贪心或者动态规划,这里增加孩子数目其rating与前面旅清还是有点困难,因此,可考虑贪心。
贪心的话,首先要保证每个孩子有1颗糖,其次,把不“和谐”的关系修理就OK。
考虑从左到右,多次修理,算法复杂度O(N*N)。
while(1){ bool te = false;for(unsigned i=1;i<m;++i){ if((ratings[i] > ratings[i-1]) && (x[i] <= x[i-1])){ x[i] = x[i-1] + 1; te = true; } else if((ratings[i] < ratings[i-1]) && (x[i] >= x[i-1])){ x[i-1] = x[i] + 1; te = true; }} if(!te) break;}一次从左至右的扫描修理,只处理了右-左正常,可能的情况是,修理之后,下一个位置会把这种合理性修改掉。如下面例子。
1 3 4 5 4 3 2 1
1 1 1 1 1 1 1 1
1 2 3 4 2 2 2 1 first process left to right
1 2 3 4 3 3 2 1 second process left to right
存在一个2-1后2-2的不合理情况,即只剩下先左后右不合理分支。处理掉这种分支,可以考虑反过来修理。(因为,反过来修理会提前把后面出现的不合理逐一转换为合理)
左右修理:(左-右)合理-合理....合理-(处理本位置使之合理,但前者就不合理了,即最后的肯定是合理的,多次循环还是可能会有合理变不合理)
右左修理:(右-左)既然最后合理,那么反过来修理,这个不合理列最右边肯定是2 1之流(连续减小才会碰到左-右修理搞不定),即最右为正确数值(value是正常),往左刷一次即可完成修理。
for(unsigned i=1;i<m;++i) { if((ratings[i] > ratings[i-1]) && (x[i] <= x[i-1])) x[i] = x[i-1] + 1; else if((ratings[i] < ratings[i-1]) && (x[i] >= x[i-1])) x[i-1] = x[i] + 1; } for (unsigned i = m -1; i>0 ;--i) { if((ratings[i] > ratings[i-1]) && (x[i] <= x[i-1])) x[i] = x[i-1] + 1; else if((ratings[i] < ratings[i-1]) && (x[i] >= x[i-1])) x[i-1] = x[i] + 1; }
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