以下是在codility.com中获得100分之100的解决方案。请阅读有关前缀和的信息，以了解解决方案：
public static int[] solveGenomicRange(String S, int[] P, int[] Q) { //used jagged array to hold the prefix sums of each A, C and G genoms //we don't need to get prefix sums of T, you will see why. int[][] genoms = new int[3][S.length()+1]; //if the char is found in the index i, then we set it to be 1 else they are 0 //3 short values are needed for this reason short a, c, g; for (int i=0; i<S.length(); i++) { a = 0; c = 0; g = 0; if ('A' == (S.charAt(i))) { a=1; } if ('C' == (S.charAt(i))) { c=1; } if ('G' == (S.charAt(i))) { g=1; } //here we calculate prefix sums. To learn what's prefix sums look at here https://codility.com/media/train/3-PrefixSums.pdf genoms[0][i+1] = genoms[0][i] + a; genoms[1][i+1] = genoms[1][i] + c; genoms[2][i+1] = genoms[2][i] + g; } int[] result = new int[P.length]; //here we go through the provided P[] and Q[] arrays as intervals for (int i=0; i<P.length; i++) { int fromIndex = P[i]; //we need to add 1 to Q[i], //because our genoms[0][0], genoms[1][0] and genoms[2][0] //have 0 values by default, look above genoms[0][i+1] = genoms[0][i] + a; int toIndex = Q[i]+1; if (genoms[0][toIndex] - genoms[0][fromIndex] > 0) { result[i] = 1; } else if (genoms[1][toIndex] - genoms[1][fromIndex] > 0) { result[i] = 2; } else if (genoms[2][toIndex] - genoms[2][fromIndex] > 0) { result[i] = 3; } else { result[i] = 4; } } return result; }
