leetcode——135.分发糖果

暴力法，多次循环检查，直到满足条件。

public int candy(int[] ratings) {
        int[] candies = new int[ratings.length];
        Arrays.fill(candies,1);
        boolean flag = true;
        int sum = 0;
        while(flag){
            flag = false;
            for(int i = 0;i<ratings.length;i++){
                if(i!=ratings.length -1 && ratings[i] >ratings[i+1] && candies[i] <= candies[i+1]){
                    candies[i] = candies[i+1] + 1;
                    flag = true;
                }
                if(i>0 && ratings[i] > ratings[i-1] && candies[i] <= candies[i-1]){
                    candies[i] = candies[i-1]+1;
                    flag = true;
                }
            }
        }
        for(int candy: candies){
            sum += candy;
        }
        return sum;
    }

 挺慢的。。。。

方法二，用两个数组，这个方法和我自己的思路比较像

public int candy(int[] ratings) {
        int sum = 0;
        int[] left2right = new int[ratings.length];
        int[] right2left = new int[ratings.length];
        Arrays.fill(left2right, 1);
        Arrays.fill(right2left, 1);
        for (int i = 1; i < ratings.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                left2right[i] = left2right[i - 1] + 1;
            }
        }
        for (int i = ratings.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                right2left[i] = right2left[i + 1] + 1;
            }
        }
        for (int i = 0; i < ratings.length; i++) {
            sum += Math.max(left2right[i], right2left[i]);
        }
        return sum;
    }

——2020.7.13