IT 討論區（48）最強工種，FG25K，你！敢唔敢成功？

JP 3xk + 日本妹
vs HK 3xk + 無日本妹
盲都識揀

其實日本p仔人工有幾多

FG應該200-300k，計返同香港一樣，但税多d

上次見到有囡上黎in
老細話太鈍唔請

今朝嗰條題目:
return int array......

所以我只可以喺香港呃飯食

BBA degree 咩都唔識
係咪就咁報個bootcamp出黎就有25k?
想轉行

-2 base
https://pastebin.com/kdwbJ8b1

唔係，淨係hr已經過唔到，25k要cs-related再加intern，埋位可以即食

唔洗報bootcamp
好易25k, HKTV果D可能未必得
因為果D真做, 實戰
我教你GOOGLE 大法, 每月逗25k

咁做唔知得唔得
將A向左sift一格, 出 B =-floor(A)
e.g. let A = 101101
B = A = 101111 => 01101

之後將floor轉ceil同乘-1, 出C = ceil(A)
C = array of zero
B[0] -=A[0]
if B[0] = -1, C[0] = 1
for i in B:
if B[i]=1: C[i]++, C[i+1]++
e.g. C = 000000 => 100000 => 111000 => 112100 => 112111

p.s. C[i] = 0, 1 or 2, |C[i]-C[i+1]|<=1

進位
for i in C:
if C[i] =2:
C[i] = 0, C[i+1] --
e.g. C = 112111 => 110011

[quote][quote][quote][quote]```
public class Solution {
	
	public int[] powerSet;
	
	public int[] solution(int[] A) {
		int length = A.length;
		if (length == 0)
			return A;
		powerSet = new int[length];
		int sum = 0;
		for (int i = 0; i < A.length; i++) {
			if (A[i] == 1) {
				int power = pow(-2, i);
				powerSet[i] = power;
				sum += power;
			}
		}
		
		int ceil;
		if (sum % 2 == 1 && sum > 0) {
			ceil = sum / 2 + 1;
		} else {
			ceil = sum / 2;
		}
		
		int index = getLeadingIndex(ceil);
		int[] array = new int[index+1];
		array[index] = 1;
		ceil -= powerSet[index];
		while (ceil != 0) {
			index = getLeadingIndex(ceil);
			array[index] = 1;
			ceil -= powerSet[index];
		}

		return array;
	}
	
	public int pow(int base, int power) {
		int number = 1;
		while (power-- > 0) {
			number *= base;
		}
		return number;
	}
	
	public int getLeadingIndex (int number) {
		int index = 0;
		int sum = 0;
		if (number > 0) {
			index = -2;
			while (sum < number) {
				index += 2;
				sum += powerSet[index];
			}
		} else if (number < 0) {
			index = -1;
			while (sum > number) {
				index += 2;
				sum += powerSet[index];
			}
		}
		return index;
	}
}

幅圖睇唔到assumption，如果length of array太大，可能要再tune[/quote]Brute force係2^n，你呢個solution係n^2，length of array如果4位數可能要再optimize[/quote]都係無野，long都係得2^63，n^2夠做[/quote]

咁做唔知得唔得
將A向左sift一格, 出 B =-floor(A)
e.g.  let A = 101101
B = A = 101111 => 01101

之後將floor轉ceil同乘-1, 出C = ceil(A)
C = array of zero
B[0] -=A[0]
if B[0] = -1:
   C[0] = 1
for i in range (len(B)):
   if B[i]=1: 
     C[i]++, C[i+1]++
e.g.  C = 000000 => 100000 => 111000 => 112100 => 112111

p.s. C[i] = 0, 1 or 2, |C[i]-C[i+1]|<=1

進位
for i in range (len(C)) :
  if C[i] =2:
     C[i] = 0, C[i+1] --
e.g. C = 112111 => 110011

[/quote]

上面個題鳩做
原理，shift 左一格再加埋一齊
shift 完再加既scenario 得4種，寫曬就得

public int[] solution(int[] A) {
    if (A.length <= 1) {
        return A;
    }
    int[] B = new int[A.length];
    for (int i = 0, j = 1; i < A.length; i++, j++) {
        if (j == A.length) {
            B[i] = A[i];
        } else if (A[i] != A[j]) {
            // 0 1    1 0
            // 1 ? or 0 ?
            B[i] = 1;
        } else if (A[i] == 0) {
            // 0 0
            // 0 ?
            B[i] = 0;
        } else {
            B[i] = 0;
            if (j + 1 == A.length || A[j + 1] == 0) {
                // 1 1
                // 1 0
                B[j] = 0;
            } else {
                // 1 1
                // 1 1
                B[j] = 1;
            }
            i++;
            j++;
        }
    }
    int effectiveIndex = 0;
    for (int i = B.length - 1; i >= 0; i--) {
        if (B[i] == 1) {
            effectiveIndex = i;
            break;
        }
    }
    return Arrays.copyOf(B, effectiveIndex + 1);
}

Online compiler
[url]http://tpcg.io/eSYcyl
[/url]

加左 test cases 係online compiler 到
time complexity：linear
space complexity：linear

其實呢唔使幫商家佬諗咁多嘢
請唔到人咪加價囉
唔係你搵咩吸引人入行
除非你係intel咁有名氣
冇人工都一堆人仆去做
賺唔賺到錢係你嘅事
賺到嘅冇問題 賺唔到咪執昱
再者好似你咁講殺出一條血路
都唔代表你有位升啦
香港啲老細

係server房同人妻扑野

識唔識Ctrl c Ctrl v ?

頂咪貼曲啦 要打J自己打啦好毒

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