484. Find Permutation

OBKoro1 发布于2019-08-14 17:28 / 3276人阅读

摘要：题目链接，用一个来指向现在到的数，每次碰到，就的数量，最后碰到把到放入结果，更新。还有一种写法是先赋值，之后检查，再。

484. Find Permutation

题目链接：
https://leetcode.com/problems...

greedy，用一个point：number来指向现在到的数，每次碰到"D"，就count"D"的数量，最后碰到I把number + count到number放入结果，更新count = 0。

public class Solution {
    public int[] findPermutation(String s) {
        int n = s.length();
        // the value in result
        int number = 1;
        int count = 0;
        int[] result = new int[n + 1];
        for(int i = 0; i <= n; i++) {
            if(i == n || s.charAt(i) == "I") {
                for(int j = i; j >= i - count; j--) result[j] = number++;
                count = 0;
            }
            else count++;
        }
        
        return result;
    }
}

还有一种写法是先赋值，之后检查"D"，再reverse。

public class Solution {
    public int[] findPermutation(String s) {
        int n = s.length();
        
        int[] result = new int[n + 1];
        // assign value
        for(int i = 0; i <= n; i++) result[i] = i + 1;
        // reverse for D
        for(int i = 0; i < n; i++) {
            if(s.charAt(i) == "D") {
                int j = i + 1;
                while(j < n && s.charAt(j) == "D") j++;
                reverse(result, i, j);
                i = j;
            }
        }
        return result;
    }
    
    private void reverse(int[] result, int i, int j) {
        while(i < j) {
            int temp = result[i];
            result[i] = result[j];
            result[j] = temp;
            i++; j--;
        }
    }
}

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