[LeetCode] 135. Candy java
2017-03-09 10:49
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/**135. Candy * @param ratings * @return */ public int candy(int[] ratings) { int len = ratings.length; if (len == 0) return 0; int[] left = new int[len]; int[] right = new int[len]; left[0] = 1; for (int i=1; i<len; i++) { if (ratings[i] > ratings[i-1]) left[i] = left[i-1]+1; else left[i] = 1; } right[len-1] = left[len-1]; for (int i=len-2; i>=0; i--) { if (ratings[i] > ratings[i+1]) right[i] = right[i-1]+1; else right[i] = 1; } int sum = 0; for (int i=0; i<len; i++) { sum += (left[i]>right[i])? left[i]: right[i]; } return sum; }
这道题和Trapping water那个是一样的想法,因为无论是水坑还是得到糖的小朋友,影响因素都不只一边,都是左右两边的最小值/最大值来决定的。
所以这道题跟上一道一样,也是左右两边遍历数组。
left数组存从左边遍历,当前小朋友对比其左边小朋友,他能拿到糖的数量;
right数组存从右边遍历,当前小朋友对比其右边小朋友,他能拿到的糖的数量。
最后针对这两个数组,每个小朋友能拿到的糖的数量就是这两个数最大的那个数,求总加和就好了。
稍稍改进一下,在遍历right时直接求和
public int candy(int[] ratings) { int len = ratings.length; if (len == 0) { return 0; } int[] nums = new int[len]; nums[0] = 1; for (int i = 1; i < len; i++) { if(ratings[i] > ratings[i-1]) { nums[i] = nums[i-1] + 1; } else { nums[i] = 1; } } int res = nums[len-1]; for (int i = len-2; i >= 0; i--) { int cur = 1; if(ratings[i]>ratings[i+1]) { cur = nums[i+1]+1; } res += Math.max(cur,nums[i]); nums[i] = cur; } return res; }
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